484 124 5MB
English Pages 271 [301]
Table of contents :
Front Cover
CHAPTER
DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS SECTION PAGE 1 Differential Equation Ordinary and Partial Order, Degree
Solution of an Equation ·
Derivation of a Differential Equation from its Primitive
General, Particular Solution
CHAPTER II
DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE 5 Exact Differential Equation Integrating Factor •
General Plan of Solution •
Condition that Equation be Exact
Exact Differential Equations II
Variables Separated or Separable
Homogeneous Equations
Equations in which M and N are Linear but not Homogeneous
Equations of the Form yfi(xy)dx + xf2(xy)dy = 0
Linear Equations of the First Order
Equations Reducible to Linear Equations
Equations of the Form x'yº (my dx+nx dy)+x°y³ (µy dx+vx dy)=0
Integrating Factors by Inspection
Other Forms for which Integrating Factors can be Found
Transformation of Variables ·
Summary
CHAPTER III
Differential Equation of a Family of Curves
Geometrical Problems involving the Solution of Differential Equations
Orthogonal Trajectories
Physical Problems giving Rise to Differential Equations
CHAPTER IV
DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE THAN THE FIRST 24 Equations Solvable for
Equations Solvable for y 26 Equations Solvable for x
Clairaut's Equation
Summary
CHAPTER V
SINGULAR SOLUTIONS 29 Envelopes 30 Singular Solutions
Discriminant
Singular Solution Obtained directly from the Differential Equation
Extraneous Loci
Summary PAGE 38 43 49 58 ii cu 52 55 56
69
CHAPTER VI
TOTAL DIFFERENTIAL EQUATIONS
Total Differential Equations
Method of Solution 76
Homogeneous Equations
Equations involving more than Three Variables
Equations which do not satisfy the Condition for Integrability
Geometrical Interpretation 84
Summary
CHAPTER VII
LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 42 General Linear Differential Equation
Linear Differential Equations with Constant Coefficients Complemen
SECTION PAGE 44 Roots of Auxiliary Equation Repeated ·
Roots of the Auxiliary Equation Complex
Properties of the Symbolic Operator (Dα)
Particular Integral
Another Method of finding the Particular Integral ΙΟΙ
Variation of Parameters •
Method of Undetermined Coefficients •
tary Function
Cauchy's Linear Equation
Summary
CHAPTER VIII
LINEAR DIFFERENTIAL EQUATIONS OF THE SECOND ORDER 53 Change of Dependent Variable
Change of Independent Variable
Summary
CHAPTER IX
MISCELLANEOUS METHODS FOR SOLVING EQUATIONS OF HIGHER ORDER THAN THE FIRST 56 General Plan of Solution ·
Dependent Variable Absent
Independent Variable Absent
Linear Equations with Particular Integral Known •
Exact Equation Integrating Factor •
Transformation of Variables •
Summary ·
CHAPTER X
SYSTEMS OF SIMULTANEOUS EQUATIONS 63 General Method of Solution •
Systems of Linear Equations with Constant Coefficients
Systems of Equations of the First Order
Geometrical Interpretation •
Systems of Total Differential Equations •
Differential Equations of Higher Order than the First Reducible to Systems of Equations of the First Order
Summary •
INTEGRATION IN SERIES
Riccati's Equation ·
Gauss's Equation Hypergeometric Series ·
Primitives involving Arbitrary Functions •
CHAPTER XIII
Nonlinear Partial Differential Equations of First Order Complete,
Special Methods
Summary
Special Method
89
SECTION PAGE
91
Nonhomogeneous Linear Equations with Constant Coefficients
Condition that a Relation exist between Two Functions of Two Variables
INDEX
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LIBRARY OF THE UNIVERSITY OF ILLINOIS
515.35 C 66e
cop:4 REMOTE STORAGE
LIDRAKT,
Return this book on or before the Latest Date stamped below. Theft, mutilation, and underlining of books are reasons for disciplinary action and may result in dismissal from the University. University of Illinois Library
OCT 19 1964
DOV 17 1964 JUL 22 1965
Aub
3 1968
JUL 10 1968
OCT 2 2 1969
MAR
3 1978
JUN
1. RECD
L161 O 1096
899

Kohuck y 1300 ·20ou
D. M. ARMSIND Beulan
Armstro
ng.
AN
ELEMENTARY TREATISE
ON
DIFFERENTIAL
EQUATIONS
BY ABRAHAM COHEN, PH.D. ASSOCIATE PROFESSOR OF MATHEMATICS JOHNS HOPKINS UNIVERSITY
D.
C.
HEATH
BOSTON
& CO. , PUBLISHERS
NEW YORK
CHICAGO
BY THE SAME AUTHOR
An Introduction to the Lie Theory of One Parameter Groups
vii + 248 pages
D. C. HEATH &
Half Leather .
CO. , PUBLISHERS
COPYRIGHT, 1906, By D. C. HEATH & CO.
IJ 6
O
A T
G E
E
515.35 свое b ca Crio 4 . . PREFACE
Crous
THE following pages are the result of a course in Differential Equations which the author has given for some years to classes comprising students intending to pursue the study of Engineering or some other Physical Science, as well as those expecting to continue the study of Pure Mathematics or Mathematical Physics. The primary object of this book is to make the student familiar with the principles and devices that will enable him to integrate most of the equations he is apt to come across. As much of the theory is given as is likely to be comprehensible to the student who has had a year's course in the Differential and Integral Calculus , and yet is sufficient to form a harmonizing setting for the numerous and otherwise apparently miscellaneous classes of equations , and the disconnected methods for solving them. It is intended to have the work sufficiently broad to make it a handy book of reference, without affecting its utility as a textbook. A number of footnotes and remarks have been put in, which, without breaking the continuity of the practical side of the subject, must prove of interest and value. Numerous historical and bibliographical references are also made. A course that is limited in point of time and aims only at acquiring skill in integrating most of the equations that are apt to arise could dispense with §§ 12 , 15 , 17 , 22 , 28 (part in small type), 33 , 34, 3840, 46–48 , 66–69 (except examples) , 70 , 71 , 73 , 75 , 78 , 80 , 81. Many of these sections should properly come in a wellbalanced course. The needs of the class, and the time at its disposal, must decide which of them , if any , should be omitted . The author has had in mind continually the necessity of systematizing the various classes of equations that can be solved by elementary means, and of minimizing the number of methods by which they can be solved . To enable the student to get a better iii
iv
PREFACE
general view of the subject, the summaries at the ends of the various chapters and the final general summary must prove of great value. Numerous applications to problems in Geometry and the Physical Sciences have been introduced , both in the body of the text and in the form of exercises for the student. Although a large number of the problems have been published before, many are new, and all have been chosen to bring out the various methods of the differential equations, and of the integral calculus as well. Many of the examples worked out in the text were chosen to recall some of the more important methods of the latter ; for while the use of tables of integrals is recommended , the student should not feel absolutely dependent upon them. Most of the solutions have a simple form or an interesting interpretation. Great care has been taken to avoid typographical errors. The author shall be very glad to learn of any that still exist. The method of undetermined coefficients for finding the particular integral in the case of linear equations with constant coefficients is believed to be presented here for the first time in its complete form . The subject of Partial Differential Equations is so vast that it was decided to present only a few topics , which, in all probability, will suffice for the needs of the students for whom this book is intended . It was only after considerable thought that the author refrained from adding a chapter on the Lie Theory. It is hoped to present that important branch of the subject in a separate volume. In conclusion the author takes great pleasure in expressing his appreciation of the valuable suggestions made by Professor F. S. Woods of the Massachusetts Institute of Technology, as well as of those by Professor L. G. Weld of the University of Iowa and Professor E. J. Townsend of the University of Illinois . ABRAHAM COHEN. JOHNS HOPKINS UNIVERSITY, BALTIMORE, MARYLAND, October, 1906.
CONTENTS CHAPTER I DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS SECTION 1. Differential Equation. Ordinary and Partial. Order, Degree · 2. Solution of an Equation 3. Derivation of a Differential Equation from its Primitive 4. General, Particular Solution .
PAGE I 2 3 5
CHAPTER II DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE 5. 6. 7. 8. 9. 10. II. 12. 13. 14. 15. 16. 17. 18. 19.
• Exact Differential Equation. Integrating Factor General Plan of Solution • Condition that Equation be Exact Exact Differential Equations Variables Separated or Separable Homogeneous Equations Equations in which M and N are Linear but not Homogeneous Equations of the Form yfi (xy) dx + xf2 (xy) dy = 0 Linear Equations of the First Order Equations Reducible to Linear Equations Equations of the Form x'yº (my dx + nx dy) + x°y³ (µy dx + vx dy) = 0 . Integrating Factors by Inspection Other Forms for which Integrating Factors can be Found · Transformation of Variables . Summary
7 9 9 II 13 14 16 17 18 20 22 23 24 26 28
CHAPTER III APPLICATIONS 20. Differential Equation of a Family of Curves . 21. Geometrical Problems involving the Solution of Differential Equations V
31 34
vi
CONTENTS
SECTION 22. Orthogonal Trajectories 23. Physical Problems giving Rise to Differential Equations
PAGE 38 43
CHAPTER IV DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE THAN THE FIRST
49 52 55 56 58
Equations Solvable for p Equations Solvable for y Equations Solvable for x Clairaut's Equation Summary
cu ii
24. 25. 26. 27. 28.
CHAPTER V SINGULAR SOLUTIONS
29. Envelopes 30. Singular Solutions 31. Discriminant 32. Singular Solution Obtained directly from the Differential Equation 33. Extraneous Loci • • 34. Summary
•
61 63 64 66 69 75
CHAPTER VI
TOTAL DIFFERENTIAL EQUATIONS 35. 36. 37. 38. 39. 40. 41.
Total Differential Equations Method of Solution Homogeneous Equations Equations involving more than Three Variables Equations which do not satisfy the Condition for Integrability Geometrical Interpretation · • Summary
76 80 81 83 84 86 87
CHAPTER VII LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS 42. General Linear Differential Equation 43. Linear Differential Equations with Constant Coefficients. Complementary Function • ·
89
91 ཌ
CONTENTS SECTION 44. Roots of Auxiliary Equation Repeated 45. Roots of the Auxiliary Equation Complex 46. Properties of the Symbolic Operator (D  α) 47. Particular Integral 48. Another Method of finding the Particular Integral 49. Variation of Parameters 50. Method of Undetermined Coefficients 51. Cauchy's Linear Equation . 52. Summary
vii 
PAGE 93 94 96 97 ΙΟΙ • 103 • 107 113 115 ·
CHAPTER VIII LINEAR DIFFERENTIAL EQUATIONS OF THE SECOND ORDER
53. Change of Dependent Variable 54. Change of Independent Variable 55. Summary
123 127 129
CHAPTER IX MISCELLANEOUS METHODS FOR SOLVING EQUATIONS OF HIGHER ORDER THAN THE FIRST
56. 57. 58. 59. 60. 61. 62.
General Plan of Solution Dependent Variable Absent Independent Variable Absent Linear Equations with Particular Integral Known Exact Equation. Integrating Factor Transformation of Variables . Summary
·
131 131 134 • 135 • 137 • 143 · 144
CHAPTER X SYSTEMS OF SIMULTANEOUS EQUATIONS • 149 General Method of Solution . Systems of Linear Equations with Constant Coefficients 150 Systems of Equations of the First Order 153 • 157 Geometrical Interpretation • 159 Systems of Total Differential Equations Differential Equations of Higher Order than the First Reducible to 160 Systems of Equations of the First Order 161 • 69. Summary
63. 64. 65. 66. 67. 68.
viii
CONTENTS
CHAPTER XI INTEGRATION IN SERIES SECTION 70. The Existence Theorem 71. Singular Solutions 72. Integration, in Series, of an Equation of the First Order 73. Riccati's Equation 74. Integration, in Series, of Equations of Higher Order than the First 75. Gauss's Equation . Hypergeometric Series .
PAGE • 164 • 168 · 169 · 173 • 177 · 192
CHAPTER XII PARTIAL DIFFERENTIAL EQUATIONS 76. Primitives involving Arbitrary Constants 77. Primitives involving Arbitrary Functions 78. Solution of a Partial Differential Equation
•
196 199 202.
CHAPTER XIII PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 79. Linear Partial Differential Equations of the First Order. Method of · Lagrange . 80. Integrating Factors of the Ordinary Differential Equation M dx+ Ndy =0 . · 81. Nonlinear Partial Differential Equations of First Order. Complete, General, Singular Solutions 82. Method of Lagrange and Charpit 83. Special Methods 84. Summary
205
209 211 215 221 227
CHAPTER XIV PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER THAN THE FIRST 85. Partial Differential Equations of the Second Order, Linear in the Second Derivatives. Monge's Method 230 86. Special Method 236 87. General Linear Partial Differential Equations 238 239 88. Homogeneous Linear Equations with Constant Coefficients
CONTENTS
SECTION 89. Roots of Auxiliary Equation Repeated 90. Roots of Auxiliary Equation Complex 91. Particular Integral 92. Nonhomogeneous Linear Equations with Constant Coefficients 93. Equations Reducible to Linear Equations with Constant Coefficients 94. Summary
ix PAGE 240 242 243 246 • 249 251
NOTES I. Condition that a Relation exist between Two Functions of Two Variables II. General Summary
253 254
ANSWERS TO EXAMPLES INDEX REFERENCES
257 269 271
DIFFERENTIAL
EQUATIONS
CHAPTER I DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS 1. Differential Equation .
Ordinary and Partial.
Order, Degree.
A differential equation is an equation involving differentials or derivatives . Thus, d2y + a²y = 0, dx2
(1)
(x dy − y)² = x² + y², dx dx dy x² +3 +3 dy = 0, y=x dx
(2)
(3)
d'y dx2
3  k = 0,
(4) I + dy dx
Əz ду
(5)
x
Əz дх
(6)
y
22% 22% + Zx dx2 dxdy
(7)
x²y³dx + x²³ydy = 0,
: 0,
მთ = xyz, dy ду
are examples of differential equations. Equations in which there is a single independent variable ( and which, therefore, involve ordinary derivatives ) are known as ordinary differential equations.
Equations ( 1 ) , (2) , (3) , (4), ( 7) are such.
If an equation involves more than one independent variable, so that partial derivatives enter, it is known as a partial differential equation.
Examples of such are equations (5), (6) . I
2
DIFFERENTIAL EQUATIONS
§2
By the order of an equation we mean the order of the highest derivative involved . Thus, equations ( 2 ) , ( 3 ) , ( 5 ) , ( 7 ) , are of the first order ; ( 1 ) , ( 4 ) , ( 6 ) , are of the second order. By the degree of an equation , we mean the degree of the highest ordered derivative entering, when the equation is rationalized and cleared of fractions with regard to all the derivatives. Thus (1 ), (5 ), (6), (7) , are of the first degree ; ( 2 ) , ( 3 ) , (4 ) , are of the second degree.
2. Solution of an Equation.
By a solution of a differential equa
tion we mean a relation connecting the dependent and independent variables which satisfies the equation . Thus y = sin ax is a soluI tion of ( 1 ) , x² + y² = is a solution of ( 4 ) , z = x + y is a solution k2 of ( 5 ) , a³³ = 1 is a solution of (7). should verify these facts. ]
[ The student, as an exercise,
Attention should be called to the fact that a differential equation has an indefinite number of solutions . It can be seen readily that y = 2 sin ax, y = 6 cos ax, y = A cos ax + Bsin ax (where A and B are any constants whatever) all satisfy equation ( 1 ) .
The student is
in the habit of adding a constant of integration when integrating a function .
He says the integral of cos x is sin x + c.
Now the prob
lem of integration studied in the Calculus is only a special case of the general problem of solving a differential equation .
To integrate
cos x dx is to find a function, sin x + c, whose derivative is cos x. Sc In the language of the Differential Equations, we should say that the dy solution of = cos x is y =sin x + c, where c is an arbitrary constant.* dx A constant in a solution will be said to be arbitrary if any value whatever may be assigned to it. Thus y = sin x + c is a solution of dy = cos x, no matter dx * It may be noted that in the special case occurring in the Integral Calculus the arbitrary constant always occurs as an additive one, while in the general case it may enter in an endless number of ways.
$3
DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
3
what value has. Similarly y = A cos ax + Bsin ax is a solution of ( 1 ) for any values of A and B. They are arbitrary constants. On the other hand, a is not arbitrary. While any value one pleases may be assigned to it in ( 1 ) , once chosen, its value is fixed, and that value alone can enter into the expression for the solution. * Restricting ourselves to ordinary differential equations, we see that a solution may involve one or more arbitrary constants. The question naturally arises, what is the maximum number of such constants a solution may contain ? 3. Derivation of a Differential Equation from its Primitive. — Just as the problem of integration is the inverse of that of differentiation, so the problem of finding the solution of a differential equation is the inverse of that of finding the differential equation which is satisfied by a relation among a set of variables, which relation may or may not involve one or more arbitrary constants . In order to make this problem precise, we shall say that we wish to find the differential equation of lowest order satisfied by this relation, and not involving any arbitrary constants . Thus y = A cos x, where A is an arbitrary
dy d²y d³y + y tan x = 0, + y = 0, y tan x = 0, dx dx3 dx2 etc. But we shall say that dy +y tan xo is the differential equation dx to which it gives rise.
constant,
satisfies
Again, y = A cos x + B sin x, where A and B are arbitrary constants, satisfies d'y + y = o, also d³y + dy = o, etc. Here , as before , dx3 dx dx2 d'y +y = o is the differential equation we are interested in. dx2 Perfectly generally, if we have a relation which involves n arbitrary constants,† we differentiate this expression n times, thus having in * The study of partial differential equations will be taken up in Chapter XII . + It is implied, of course , that the n constants are essential ; that is, that they cannot be replaced by a smaller number. For example, y = x + a + b really involves only one essential constant, since a + b is no more than a single constant. Again aer+b is no more general than aex.
DIFFERENTIAL EQUATIONS
4
$3
all n + 1 equations from which to eliminate the n constants.
So that
a relation (or, as we shall henceforth call it, a primitive) in which n arbitrary constants appear, gives rise to a differential equation which involves derivatives of as high an order as the nth . This process is unambiguous ; hence a primitive gives rise to one and only one differential equation .
Without going into a rigorous proof of the
fact, we see how a primitive involving n arbitrary constants gives rise to a differential equation of the nth order. To illustrate, find the differential equations corresponding to the following primitives : Ex. 1. y = Ge * + dy Then = αce dx
*. Here ₁ and c₂ are the arbitrary constants.
+ α«€¿Cª2*,
d'y = dx2 From these three equations we must eliminate
1 and
2.
Con
sidering them as three homogeneous equations in the quantities 1 , ₁₁*, ₂eª½*, we have
I,
I
ala
α19
α2
d'y dx2'
α12,
ولا dy dx'
= 0, or day  (a (α₁ + α₂)) d² + α₁α₂y = 0. + αs dx dx2 α2
Ex. 2. (x − c)² + y² = r².
Here
is the arbitrary constant.
Then
dy From these two equations we must = 0. x c + y dx eliminate c. dy x  c = —・y・ • Substituting this in the original equation, Now dx we have 22 1º (dx 2x) ² +y² = r².
$4
DIFFERENTIAL EQUATIONS AND THEIR SOLUTIONS
5
Ex. 3. y = cx + √1 − c².
Ex. 4. (x − c₁)² + (y − c₂)² = r². Ex. 5. y =
x² + Cq
Ex. 6. y² + c₁x = 0. Ex. 7. x = 2 cy + c². 4. General, Particular Solution .
If now we start with a differ
ential equation, its solution involving the maximum number of arbitrary constants is nothing but the primitive which gives rise to the differential equation. That solution cannot contain more than n arbitrary constants, by the theorem in § 3. Besides, it must contain as many as n ; otherwise it would be the primitive of a lower ordered equation. The solution involving the maximum number of arbitrary constants is called the general (or complete) solution. *
By means of the
general existence theorem (§ 70), we can prove the following theorem :  The general solution of an ordinary differential equation of the nth order is one that involves n arbitrary constants. Attention should be called to the fact that although the general solution may assume a variety of forms, all of these give the same relation among the variables, so that there is actually only one general solution ; thus it is readily seen that x313 = C is a solution of (7) ; so also is log x + log y = C, or log xy = C. These, obviously, are all equivalent to saying that xy is constant. The uniqueness of the general solution is part of the existence theorem. A solution which is derivable from the general solution by assigning fixed values to the arbitrary constants is called a particular solution. Thus, y = cos x and y = cos x sin x are particular solu
tions of d'y + y = 0 . dx2 * We shall see later that there may be solutions which are distinct from the general solution. In the general theory of Differential Equations the existence of a solution for every differential equation (under certain restrictions) is proved. The solution there obtained is the general solution referred to in the text.
DIFFERENTIAL EQUATIONS
6
$4
As mentioned in § 2, the problem of the Differential Equations includes that of the Integral Calculus as a special case. Thus, in the latter the general problem is to solve dy f(x). dx This is only a special case of the problem of finding the solution of the differential equation of the first order involving two variables,
dy dx =f(x, y) where ƒ(x, y) may be a function of both the variables . We speak of integrating or solving the equation, in the general case, and at times refer to the simpler problem of the Integral Calculus as performing a quadrature. A function of the independent variable will be said to be an integral of the equation if, on equating it to the dependent variable , we have a solution.
We have a general or particular integral accord
ing as the resulting solution is general or particular . While the problem of finding the differential equation corresponding to a given primitive is a direct one, and can be carried out according to a general plan, involving simply differentiation and elimination, that of finding the primitive or general solution of a given differential equation, like most inverse problems, cannot be solved by any general method. In the following chapters we shall bring out, in as systematic a manner as possible, some of the classes of equations whose solutions can be found . We shall understand that the problem of the Differential Equations is solved when we have reduced it to one of quadratures, that is, to a mere process of the Integral Calculus. While in the general theory of the Calculus it is proved that every function has an integral, it may not be possible to express it. In such cases we shall content ourselves by simply indicating this final process.
CHAPTER II DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND THE FIRST DEGREE Integrating Factor.
5. Exact Differential Equation .
The general
type of an equation of the first order and degree is (1)
M dx + N dy = 0,
where M and N are functions of x and y. Making use of the theorem that every differential equation has a general solution ( § 70), this equation has a solution containing one arbitrary constant. (2)
Solving for the constant, the solution has the form u (x, y) = C.
The differential equation having this primitive is obviously
du du ・dx +· dy = = 0. дх ду Since this must be the same equation as ( 1 ), we must have the corresponding coefficients proportional, i.e.
du Эх M
ди dy N.
If we call this common ratio μ, (which is, at most, a function of x and y) , we have
du ax So that
du
µM,
= μΝ. ду
(Mdx + Ndy) = du. 7
DIFFERENTIAL EQUATIONS
8
$5
We shall speak of an expression which is the differential of a function of one or more variables as an exact differential. Thus, (Mdx + Ndy) is such, since it is the differential of u. We shall further speak of a differential equation as an exact differential equation, if, when all the terms in it are brought to one side, that member is an exact differential. The above result can now be stated as follows : Assuming the existence of the general solution of the differential equation (1 ) , a factor µ (x, y) exists which, when introduced, will make the equation exact. This factor is known as an integrating factor, because, as we shall see (§ 8) , when our equation is exact, its integration can be effected readily. A differential equation of the first order and degree has an indefinite number of integrating factors. Suppose μ to be an integrating factor. Then µ (Mdx + Ndy) = du (x, y) Ju and μN = Ju = where the sign of identity means that μM Эх dy Now, if (u) is any continuous function of u, we have Ju Ju dx + $( u) µp(u)M dx + µp(u) N dy = p(u) dy = d↓ (u) = d (x, y), Эх ду where
dy(u) = du = 4 ( u ) , or ¥ ( u) = Sp ( u) du.
Hence up(u) is also an integrating factor. Since ( u) may be chosen in an indefinite number of ways, we see that the number of integrating factors is infinite. [For another proof, see Ex., § 7 ; also § 80. ] Thus, it is obvious by I inspection that x dy  y dx = 0 has 2 for an integrating factor. We have, actux Here μ = I " u = 1. Then I & will be an inally, xdy= x2ydx= d ("). 2 +(2/2) x +x+ + I Xx I or tegrating factor. In particular, is an integrating factor giving x² y xy I x2 or I dy dx which is log( log ) . Similarly, x y y2 gives d( − 3) .
$$ 67
THE FIRST ORDER AND THE FIRST DEGREE
9
6. General Plan of Solution.
Since every differential equation of the first order and degree which can be solved by elementary means has integrating factors, it would seem natural to try to find such a factor when the problem of solving an equation of this type arises. Practically, this is not always possible or desirable .
In the following
paragraphs of this chapter will be found the more important and the more frequently occurring classes of equations of the first order and degree which can be solved by elementary means ; and it will be noticed that they will be solved, in general, either by finding integrating factors for them or by transforming them into other forms for which integrating factors are known . 7. Condition that Equation be Exact. 
If the equation is exact
to begin with, of course, no integrating factor need be sought.
We
must, then, find the necessary and sufficient condition for exactness of an equation. If Mdx + Ndy = 0 (1) is exact, that is, if M dx + N dy is du du = N, and = M and then дх ду әм (2) ду 22u 22μ * since (2) is, then, дх ду дудх of the equation.
the differential of some function u,
ƏN дх a necessary condition for exactness
We shall now prove that it is also sufficient.
Even more, we shall
show that if ( 2) holds, we can actually find a function u such that its differential is Mdx + N dy, or, what is the same thing, such that Ju du = M, and = N. (3) дх ду
әм * Assuming the continuity of M and N, and the existence and continuity of JN . ду and Эх
ΙΟ
DIFFERENTIAL EQUATIONS
$7
In order that the first of these relations ( 3 ) should hold, we must have
u= =S* Mdx + Y(y) *
(4)
where Y is a function of y only, and plays the rôle of a constant of integration, since y is considered a constant in the process of integration
involved in (4 ) .
The value of u given by (4 ) will
satisfy the first of ( 3 ) , no matter what function of y Y may be.
In
order that u satisfy the second of ( 3) we must have
du ду
dy = = მ N; By S Mdx + ду dy
that is, y must satisfy the equation dy
(5)
x a N ― by S Mdx.
dy Since the lefthand member of ( 5 ) is a function of y only, the same must be true of the righthand member ; that is, the latter must be free of x, or in other words it must be a constant as far as x is concerned, and its derivative with respect to x must be zero . ƏN ƏM As a matter of fact, that derivative is ,t which is zero dy дх because of (2 ) .
We can, then, find Yto satisfy (5 ) , viz. a Y= S [ N = ду 3 ,SMdx] dy,
and the resulting value of u in ( 4 ) will satisfy both equations ( 3) . Ex.
Using the fact that ( 2) is the necessary and sufficient con
dition for exactness , prove that if u is an integrating factor, such that μ(Mdx + Ndy) = du , µ þ ( u) is also an integrating factor. *
By STMdx we mean the result of integrating Mdx considering y as a constant. a Mdx = M, since, in both of the processes involved, y is con† Obviously dx=S S&M sidered constant.
THE FIRST ORDER AND THE FIRST DEGREE
$8
II
8. Exact Differential Equations . This suggests a method of solving an exact equation .
For, in this case, Mdx + Ndy = du where x  a * Mdx] dy. 3,§ u = S Mdx + S [ N = dy.
So that the general solution is
a = C. ,SM dx]dy S" Mdx + S [ N = 3 dx dy = c.
(6)
Expressed in words, the operations involved in (6) are : Integrate Mdx, considering'y as a constant, thus obtaining §*Mdx. Subtract ing the derivative of this, with respect to y, from N, a function ofy
Μ only is left. The integral of this function plus §*Mdx is the lefthand member of(6). a Remark. — Frequently NM dx is nothing but those terms ду. S of Nfree of x. This suggests the simple rule : Integrate Mdx, considering y as a constant ; then integrate those terms in N dy free of x. The sum of these equated to an arbitrary constant is the general solution. This rule may fail because SMdx is not unique as far as terms involving y only are concerned. either
x² + xy or
(x + y)².
Thus (* (x + y)dx may be
See also Ex. 3.
But the rule is found
to work so often that it seems worth mentioning, with the understanding, however, that when it is employed , the result be redifferentiated to see whether the original equation is obtained . As an exercise let the student show that the general solution may also be obtained in the form a Эх S'Ndy]dx =c. S'Ndy +S [ M− 3
2 xy + I dx + Y x dy = 0. 12 y a (y ―I (2xy+ 3 1) = dx y² ду y2
Ex. 1.
x dx = x² + S* 2xyy+ 1 dx = y
; hence equation is exact.
I is the only term in N free of x. y
12
DIFFERENTIAL EQUATIONS
$8
Stdy = logy. x
.. x² +
+ logy = c is the general solution.
y 122 x2
dx +
Ex. 2.
xy²  x3
212  x2 dy = 0 . y3  x²y
Let the student prove that this is exact by showing that ƏM IN = ду Эх
x2 (y² — x²) — x² x =S S xy² 2x³ dx = Sd x(y² x²)
x2 S73dx =
log x + log(y²x²). 2 At first sight it might seem that no term of N is free of x. But I y writing it in the form ² +(y²x²) = __ + " we see that there y² —x² y y(y²x²) is such a term, viz . 1 . y .. the solution is log x + log ( 22) + log y = c, or x²y² (y² — x² ) = c.
dx Ex. 3.
+ √x² + y²
I x Gx√x² +y²=) dy = 0 .
Let the student prove that this is exact. 20 dx = log (x + √x² + y ) , the form given in integral S √x² + y² tables. I which is free of x. Hence if the rule is N contains the term y followed blindly, the solution would seem to be
log (x + √x² + y²) + log y = c.
$9
THE FIRST ORDER AND THE FIRST DEGREE
13
As a matter of fact, the solution is log (x + √x² + 1º) = C. dx x The trouble here is with the value of S √x² + y²2 dx dx a x 2 = = log x + Samaa)² + 1) a √(₁ x 2 √x² + a² S +I a
= log (x + √x² + a²) — log a. Naturally in tables of integrals the term loga is omitted, as that may be incorporated in the constant of integration. In the case of the problem under discussion it makes a big difference whether this term is used or not. If it is used, the rule gives the correct result. But if the form of the integral as given in the tables is used, then the rule is at fault. Hence the caution given in the remark above should be heeded .
Ex. 4. ( y + x) dx + x dy = 0 . Ex. 5. (6x  2y + 1 ) dx + ( 2y — 2x — 3 ) dy = 0 . 9. Variables Separated or Separable. In case M is a function of x әм ON = only, and None of y only, the relation is obviously satisfied. ду дх In this case we shall say the variables are separated. is, of course,
The integral
SMdx +SNdy = c. Very frequently, the variables are separable by inspection.
Ex. 1. sec x cos²y dx
cos x sin y dy = 0. sin y Hence dy = 0, cos²y or sec²x dx  tan y sec y dy = 0. The general solution is tan x ― secy = c. I *Here is an integrating factor, found by inspection. cos x cos²y sec x dx COS X
DIFFERENTIAL EQUATIONS
14
§ 10
Ex. 2. (1 + x)y² dx  x³ dy = 0. Ex. 3. 2 (1  y ) xy dx + ( 1 + x²) ( 1 +y²) dy = 0.
Ex. 4. sin x cos² y dx + cos² x dy = 0. 10. Homogeneous Equations. A widely occurring class of equations where the variables can be separated, not by inspection, but by a simple transformation of variables, is that in which M and N are homogeneous functions of x and y, and of the same degree . Then M is a homogeneous function of degree zero, and is therefore a funcN tion of 2* . x Our equation can now be written M = N F (2)
dy dx
Let y be a new variable, say v . x
Then y = vx,
do dy = = v+ x = F(v) , dx dx dx
dv or
=
1
F (v) v
x
and our variables are separated. Integrating this, and then replacing v by its value in terms of x and y, we have the general solution. *A convenient definition of a homogeneous function of x and y of degree r is , that if in the function we replace x and y by tx and ty respectively, where t is anything we please, the result will be the original function multiplied by tr. (This definition can be extended at once to a function of any number of variables. It is obviously consistent with the old definition of homogeneity of polynominals.) This definition can be formulated thus : Ifƒ (x, y) is a homogeneous function of degree r, then ƒ (tx, ty) = trƒ(x, y).I I When If we put t = " we have ƒ ( 1,2 ) =1 , ƒ (x, y) , or ƒ(x,y) = xrf x xr x'ƒ( 1,1) .
r= o, we have f (x, y)) = ƒ ( 1, 2 ) = F ( 2 ).
THE FIRST ORDER AND THE FIRST DEGREE § 10 Ex. 1. ( xe² +y)dx + y) dx − – x dy = 0. Then
Put y = vx, dy = v dx + x dv.
x(e" + v) dx  xv dx  x² dv = 0,
dx
dv = 0.
or
Integrating , we have
x y
= c, or log x + e
log x + e
= C.
dy = 0. − ( x³ + 2 xy³) dx /xx 1 Ex. 2. 2 x²y + 3 µ³ — dv dy Put y = vx, dx = 2 + xdx •
I + 2 v2 dv 0+ 23
dx = 0, x
v dv dv + ข I + v2
or
log v +
Then
dx = o. x
Integrating, we have
log ( 1 + v²) — log x = k,
or
log v² + log ( 1 + v²) — log x² = 2k,
or
v² (1 + v²) e2 = c ; whence x2
y² (x² + y²) = cx®. Ex. 3. (y²xy) dx + x² dy = 0.
dy Ex. 4. 2x²y + y³ — x³.dx = 0.
By $ 17
Ex. 5. y³dx + x³dy = 0 .
Ex. 6. (x +y cos ) dx
 x cos2dy = 0 . X
15
DIFFERENTIAL EQUATIONS
16
SII
11. Equations in which M and N are Linear but not Homogeneous . If M and N are both of the first degree but are not homogeneous , we can, by a very simple transformation, make them homogeneous. Suppose our equation to be of the form (a₁x + b₁y + c₁) dx + (ax + b₂y + c₂) dy = 0. Put x = x + a, y = y' + B.
The equation becomes
(a₁x' + b₁y' + a₁α + b₁ẞ + c₁) dx' + (a₂x' + b₂y' + aα + b₂ß + c₂) dy = 0. Now we may choose a and ẞ such that α₁α + b₁ß + c1 = 0, and
i.e. if we put
aα + b²ß + c₁ = 0 ;
α=
― b₁c  bc and ẞ = = c1a2 C2a1" " a1b2 a₁b₂ ab₁ a₁b₂ab₁
our equation takes the form (a₁x' + b₁y') dx' + ( aqx ' + b₂y') dy' = 0, where the coefficients are homogeneous.
Ex. 1. (4x + 3 y + 1 ) dx + (x + y + 1 ) dy = 0 . Putting xx' + 2 , y = y' ―  3, this becomes (4 x + 3y') dx' + (x ' + y') dy' = 0.
Now, ify' = vx ' , we get
dx' = I+V dv + 0, 4 + 40 + 0² dv
dv
2+v
(2 + 2)²
or
+
dx' x'
I ..log (2 +2) +
+ log x ' = c,
2 +v or
log x' (2+ ) = c⋅W
I
" whence, 2 +v x'
log (2x + y ) = c
2 x' + y'
§ 12
THE FIRST ORDER AND THE FIRST DEGREE
17
Passing back to x and y, this becomes X 2 log (2x +y  1 ) = c —
2x + y = 1 Ex. 2.
(4x  y + 2 ) dx + (x + y + 3) dy = 0.
Remark. This method breaks down in case a1 b2  a2 b₁ = 0. But in this case we can find a transformation which will separate the variables at once. For we have 2 ხა = k, a constant, and the equation takes the form a1 b1 (ax + by + c1) dx + [ k ( a1x + b₁y) + c2] dy = 0. If now, we put a₁x + b₁y = t, so that y = tax , our equation takes the form b1
(t + cs) dx + (kl + c₂) ( dt −b1 = 0, a1 dx) =
dx +
or
kt + c2 dt = 0, (bi  a₁k) t + bici  a1c2
where the variables are separated. Ex. 3.
(2x + y) dx
Į
(4x + 2y  1) dy = 0. +
12. Equations of the Form yfi (xy) dx + xf₂(xy) dy = o.
Another
class of equations in which the variables can be separated by a simple transformation may be mentioned . equation is
The general type of such an
yf₁ (xy) dx + xf2 (xy) dy = 0, where byf(xy) we mean a function of the product xy. v ข If we put xy = v, or y == 2 , then x dy = dv ― dx, and our equax x ข ข tion becomes  fi (v) dx +f2 (v) dv, x x f₂ (v) dx = 0, or
f2 dv
( /i  J/ ) where the variables are separated .
+
dx = X
18 . Ex. 1.
DIFFERENTIAL EQUATIONS
§ 13
(y + 2xy² — x²y³) dx + 2x²y dy = 0 .
x dv  v dx = Putting y = =7 , our equation becomes x2 x, dy
—· ( 1 + 2 v − v²) dx + 2 xv x or
dx 2 dv + = 0. I  72 x
Integrating, we get, since
310
log
or
x dv  v dx = 0, x2
I 2 = + I  v2 I+ V
I 
I+ V = k, + log x = I V
x I + V = C. I V Replacing v by its value, we get
x + x²y = C. I  xy Ex. 2. (2y + 3 xy²) dx + (x + 2 x²y) dy = 0.
Ex. 3. (y + xy²) dx + (x − x²y) dy = 0. 13. Linear Equations of the First Order.
A linear differential equa
tion (of any order) is a special kind of equation of the first degree. It is not only of the first degree in the highest derivative, but it is of the first degree in the dependent variable and all its derivatives. Thus, the general type of a linear differential equation of the first order is dy + Py = Q dx
where P and Q are any functions of x only.
$ 13
THE FIRST ORDER AND THE FIRST DEGREE
19
An integrating factor for this equation is readily seen to be d elPaz, since · (yeSpax) = = eSPax(dy Py dx dx + P ) . Introducing this factor, we have the solution by means of a single quadrature, thus
YesPax
ef dx + c.t = SQ eSpax
Ex. 1. dy +y cot x = secx. dx Here elPdx eſcot xdx = elog sinx =
sin x is an integrating factor.
Using this we have
sin x sec x dx = tan x dx = log cos x + c. y sin x = = Star = Ssin
Hence the solution is
y sin x = ― log cos x + c ,
or
y sin x = log sec x + c.
Ex. 2. x² dx + ( 1 + x) y = e . * We shall see later (§ 80) that this form of integrating factor arises by a perfectly general method. Excepting for pedagogic reasons, that method could be given at this point, without assuming anything to prevent our using it in this connection. + While the above method of solving a linear differential equation of the first order is undoubtedly the simplest in both theory and practice, and besides has the advantage of being readily retained in mind , it may be well to call attention to a second method , which is part of a general method applicable to linear equations of any order (see §§ 53, 59). dy dv dv + dy • Let y = y10. Then dx = y1 dx dx v ; and the equation becomes 1x + dy1 dy1 + Py₁ = o, we get y₁ = e SPdx, and dx + Py₁v 11)v = Q. If we choose y₁ so that dx dv dx dx + c; and we have the equation in v becomes dx QefPax, whence v = Se espaz eSpac dx + ce SPås, finally y = eSpd:SQ eJ eSPdx
20
DIFFERENTIAL EQUATIONS
§ 14
t
Putting this in the proper form (where the coefficient of dy is dx unity) , we get dy dx ++ x (1 + 1) = = ==
SPdx  Sdx + dx) = logx + x.
.. an integrating factor is e log +x, or x ex.
I +6. Using this, we get yxe" =Se² dx = 12 e²² + Ex. 3. dy dx Ex . 4.
2y = (x + 1)³. x+ I + 4x²y = 2 .
(x + 2 )
dx Ex. 5. x2dy + ( 1 − 2 x ) y = x². dx 14. Equations Reducible to Linear Equations.  At times an obvious transformation will change an equation into a linear one. Such a one, of frequent occurrence, is
dy + Py = Qyn * dx where, as before, P and Q are functions of x only, and n is any number.
Dividing by y" we get y ndy + Py  n+1 = Q. dx I dv dy +1 and our equation becomes Let yn+¹v, then y"! = dx In dx' dv dx + ( 1 − n) Pv = (1 − n) Q, which is linear.
* This is known as Bernoulli's Equation, after James Bernoulli ( 16541705) .
§ 14
THE FIRST ORDER AND THE FIRST DEGREE
21
dy Ex. 1. (12 ) dx 2 (1 + x) y = y§ . I  I
Lety
I +x ½ dy 2 y I  •23) 2 = dx 2 y§ dy dv en th = v, dx dx 2
dv I I +x V == 3 +3 ― dx I x3 21
··
dx I + 2 x dx 3 + 3x dx = 2 S₁ ²x + S x + x + x² · x I² S3 = 2 log ( 1  x) + log ( 1 + x + x²) . 1+x+x2 I + x + x² e log (1x)2 , or .. an integrating factor is (1 − x)2 and we have
ข
1 + x + x²  x)² (1 −
y = 1 + x + x²
3 2
I I  x3
3 2
(1 − x) 3 dx
1 + x + x² dx  x)² (1 −
I
3 I − x) 2 + c ; 4 (1 
I 3 + c, 4 (1 − x) ² +C,
whence (1  x)2
I =3 + c (1  x)² 4 1 + x + x² I + x + x² °
or
As examples of other cases where an obvious substitution will transform the equation into a linear one, consider the following :
Ex. 2. ydy + x² = x. dx
[ Put y² = v. ]
Ex. 3. sin y dy + sinx cos y dx
sin x.
Ex . 4. 4 x dy + 3y + e*x¹y = 0. dx y +1 dy Ex. 5. Vy + I. dx x + I
[ Put cos y = v. ]
l.
DIFFERENTIAL EQUATIONS
22
§ 15
15. Equations of the Form x'y (my dx + nx dy) + xºy° (µy dx + vx dy) = 0. Since d (x
) = xa ¹yb 1 (ay dx + bx dy) , it is easily seen that if
we start with any expression x'y (my dx + nx dy), we can make it exact by multiplying it by xay , provided
= B + s + I, cm = a + r + 1 , and cn = where C is any number.
As a matter of fact
xem  yen o  ngay (mydx + nx dy ) = xem lyn  1 (my da + nảy ) I = d(xemzen). C If co, this term must be replaced by d log xy". The object of introducing the undetermined quantity is to enable us to find an integrating factor for an equation of the form xy (mydc + nảy) +
gia luy dọc + vảy )= o.
r 1yens1 will render the first set of terms Just as the factor xemrly exact, so xyμlyyvo1 will render the second set of terms exact . In order to find an integrating factor for the equation, we must so determine c and1 γ that these two factors are one and the same, i.e.
we must have cm —r— 1 = yμ — p — I , cn  s I= yv  σ
I.
These two equations are sufficient, in general, to determine c and
*
Ex. 1. xy (3y dx + 2 x dy) + x² (4 y dx + 3 x dy) = 0. Here,
m = 3, n = 2, r = 4 , s = 1 , μ = 4, v = 3, p = 2 , σ = 0. 3c5 =4Y  3,
2C  2 = 3Y  I .
From these we have, c = 2 , y = 1. Hence the integrating factor is xy . Introducing this, we get the solution xy + x+y³ = C1, or x64 + 2 x¹³
k.
* If my = nμ, the equation reduces to the simple form my dx·— nx dy = 0.
§ 16
THE FIRST ORDER AND THE FIRST DEGREE
23
Ex. 2. y² (3y dx — 6 x dy) — x (y dx — 2 x dy) = 0. Ex. 3. (2 x³y — y²) dx − (2 x² + xy) dy = 0. 16. Integrating Factors by Inspection . — Integrating factors can frequently be found by inspection on closely examining the terms entering in the equation. *
Of course no general rule for this can
— y dx. A commonly occurring group of terms is xdy  y dx or x dy  y dx or dy dx or x dy  y dx dy x This suggests , " " x2 x y 12 I±
be given.
I
I
I
I
the factors being respectively
So that
is an x²
y²' xy' x² ± ²²
integrating factor for an expression of the form x dy — y dx +f(x) dx, I while may be used for one of the form x dy  y dx +f(y) dy, and y2 I for x dy  y dx 1 for x dy  y dx +f(xy) (x dy + y dx) , and + y² x² xy +f(x² ± y²) (x dx ± ydy) . Other combinations will occur to one in
actual practice. (Ex. 3 , § 10.)
Ex. 1. (y²  xy) dx + x² dy = 0.
Writing this y dx  x(y dx  x dy) = 0, I we see that is an integrating factor . Using this, we get xy2 dx x
whence
y dx  x dy
=0; y²
log x  x = c. y
Ex. 2. x dy  y dx = x dy. √x²  y² * An illustration of this we had in § 9, where by the introduction of a factor the variables were separated and the equation thus rendered exact.
Writing this
817
DIFFERENTIAL EQUATIONS
24
x dy  y dx = = xdy, we see at once that
is an x
x
I
x integrating factor.
Using this, we get
x dy y dx = dy ; whence I x2 x
sin12 = y + c. x Ex. 3. (x + y) dx — (x − y) dy = 0. Writing this x dx + y dy + y dx  x dy = o, we see at once that I is an integrating factor. x² + y²
Ex. 4. (x² + y²) dx  2xy dy = 0. Ex. 5. (xy) dx + 2 xy dy = 0. Ex. 6. x dy  y dx = (x² + y²) dx.
17. Other Forms for which Integrating Factors can be Found. ƏM applying the test for exactness ( § 7 ) , we find the value of ду equation If not, it is exact. the zero, to be If this turns out contain either Mor N as a factor.
In
ᎧᎳ дх may
By the general method of § 80
(already referred to in a footnote, § 13 ) , it will be seen that if ƏM ON дх dy is a function of x only, say f (x) , then ef (x)dx is an inteN. ᎧᎳ әм дх ду grating factor, and if is a function of y only, say f2 (y) , M then eſ½( )dy is an integrating factor.
$ 17
THE FIRST ORDER AND THE FIRST DEGREE
25
It may be interesting to call attention to the fact that this method of § 80 also I informs us that is an integrating factor in case M and N are homogexM + yN I is an integrating factor if neous and of the same degree, and that xM  yN M = yfi(xy) , N = xf2(xy) .* · These two classes of equations were considered in §§ 10 and 12, where we found transformations that separated the variables. Leibnitz ( 16461716 ) and his school endeavored to solve all equations of the first order and degree by introducing new variables which become separable in the transformed equation. Euler ( 17071783) and his school tried to solve all equations of the first order and degree by finding integrating factors. As a matter of fact, these are frequently spoken of as Euler factors or multipliers. But the idea of an integrating factor seems to be due to a contemporary of his, Clairaut ( 17131765 ) . Now it is interesting to note that just as every differential equation of the first order and degree has an integrating factor, in general, so it can be proved † that by a proper change of variables every such equation can be transformed into one in which the variables are separable. But in actual practice it would be awkward and difficult to carry out this method in all cases, just as it would be inadvisable to insist upon finding an integrating factor in every instance. These two classes of equations are of particular interest as affording examples of cases where both the general methods of solution can be readily applied.
2 Ex. 1. (3x² + 6xy + 3y²) dx + (2x² + 3 xy) dy = 0 . ƏM ƏN ax I dy 2 = x is an integrating factor. x N Ex. 2. 2 xác + (x tr +2 y ) dy = 0. ƏN ƏM дх ду = 1. .. eſay = e” is an integrating factor. M * These methods cease to apply if xM +yN= o in the first case, and if xM yN=o in the second. But in either of these cases, the solution of the equation is effected dy =  M directly with ease. For if xM +yNo, the equation takes the form dx N x and its solution is y c; on the other hand, when xM  yNo, the equation takes x M the form dy 2, and the solution is xy = c. dx N x + See Lie, Differentialgleichungen, Chapter 6, § 5 ; also the author's Lie Theory, § 20.
DIFFERENTIAL EQUATIONS
26
§ 18
Ex. 3. (y² + 2y) dx + (xy³ + 2 yª — 4 x) dy = 0. − x¹) dy = 0 . Ex. 4. (xy  y¹) dx + (y³x Ex. 5. Solve examples of §§ 10 and 12 by the method of this
paragraph. Ex. 6. (²x² + 2 mxy) dx + (my²  mx²  2 xy) dyo. 18. Transformation of Variables.  In case the equation to be integrated does not come under any of the heads treated in this chapter, it is possible, at times, to reduce it to one of them by a transformation.
No general rule for doing this can be formulated .
The form of the equation must suggest the transformation to be tried.
The following examples will illustrate .
Ex. 1. x dy  y dx + 2 x²y dx  x³ dx:= 0.  y dx suggests the transformation y = v. Here x dy — x transformation, our equation reduces to
Making this
dv + 2 xv = x, which is linear. dx An integrating factor is el2zdx or ex¹. I : . ve" = √xe* ' + c. Sxe" dx = '2 c e² Hence the general solution is I y୯୫୫ Le x = te + a
Ex. 2. (x +y) dy
dx = 0.
Here x +y suggests the transformation x + y = v. our equation reduces to v dv− (v + 1 )dx = 0,
Making this,
in which the variables are separable at once, so that we have v dv dx = dv v+ 1
dv
• V+I
$ 18
THE FIRST ORDER AND THE FIRST DEGREE
27
Integrating, we have x = vlog (v + 1 ) + c,
i.e.
x= x +y  log (x + y + 1 ) + c,
or
log (x + y + 1 ) = y + c.
This can also be integrated as follows : Writing it in the form dx ― x = y, dy we see it is linear, considering y as the independent variable. eSay = e y. An integrating factor is
xe  y = =Syevdy = −ye  v − e− + C,
Hence or
x + y + 1 = cey.
Ex. 3. x dx +y dy + y dx  x dy = 0.
(Ex. 3, § 16.)
·d Here x dx +y dy suggests x² + y², while y dx − x dy suggests
= tan combination suggests the transformation x² + y² = r², y X what is the same thing, xr, cos 0, y = r sin 0.
Then
x dx +y dy= rdr,
ydxx dy— r² do, dx = cos 0 dr— r sin 0 do, dysin 0 dr + r cos 0 d0. Our equation then becomes dr r
· de = o,
and the solution is
/ ४
=
y log r — 0 = c, or log √x² +y² — tan−¹
. This
; or,
28
DIFFERENTIAL EQUATIONS
Ex. 4. x
dy  ay + by² = cx2a. * dx
§ 19
This special form is characterized
by the fact that, when the first term is x
dy " the coefficient of y is the dx'
negative of half the exponent of x in the righthand member. Putting y = xv, the equation becomes
269+1 dv + bx2a72 = cx²ª, dx dv = dx c  bv² x¹a'
or
in which the variables are separated. 19. Summary.  In actual practice, when the equation Mdx + Ndyo is to be integrated , we proceed as follows :
By inspection we can tell when 1° the variables are separated or readily separable (§ 9), 2° M and N are homogeneous and of the same degree (§§ 10, 17 ) , 3° the equation is linear or directly reducible to one that is ( §§ 13 , 14) , 4° M and N are linear but not homogeneous (§ 11), 5° M = yfı (xy) , N = xƒ½ (xy) (§§ 12 , 17 ) , 6° the equation is of the form x'ys (my dx + nx dy) + xyº (µy dx + vx dy) = 0 (§ 15 ) . If, on inspection (and with a little practice this inspection can be made very rapidly) , the equation does not come under any of these heads, apply the test for an exact differential equation . † happen that
* This is a special form of Riccati's equation (see §73) . + It may be possible to recognize by inspection that an equation is exact. In such case proceed at once to integrate. Or an integrating factor may be obvious by inspection (see 10° below) . The general plan is to recognize, as promptly as possible, the general head under which any particular equation comes. In this summary and those of succeeding chapters the various possible methods are arranged in the order of the ease of application of the test as to whether any particular method applies.
THE FIRST ORDER AND THE FIRST DEGREE
$ 19 әм 72 ду
ду ƏN дх
әм ƏM = Mf₂y) (§ 17) . ду
8°
9°
ᎧᎳ дх = 0 (§ 8) , IN дх = Nf (x) (§ 17),
ƏM
29
If none of these cases arise, it may be possible to
10° find an integrating factor by inspection (§16) , or to 11° find some transformation that will bring the equation under one of the above heads ( § 18) . 12° As a final resort the methods of §§ 80, 25 , and 72 may be tried. Ex.
1.
I  y² dx + y √1  x2 dy = 0. x√1
Ex.
2.
I − x² dy = 0 . √1  y² dx + √1
Ex . 3.
dy 1/2 — x²y = x³ . dx
Ex.
4.
dy (y  x) ² dx == 11..
Ex.
5.
dy x² dx + y + xyz = 0.
[ Put y  x = v] .
Ex. 6.
( 1  x) y dx + ( 1 − y) x dy = 0.
Ex.
(y  x) dy + y dx = 0 .
7.
Ex. 8.
xdy  y dx = √x² + y² dx.
dy Ex.
9.
Ex. 10.
(x + a)
3y = 2 (x + a)³.
x dy  y dx = √x² — y² dx.
Ex . 11 . (x sin
y  y cos² ) dx + x cos x dy = o
Ex. 12. (x  2y + 5) dx + ( 2x − y + 4) dy = 0.
DIFFERENTIAL EQUATIONS
30
Ex . 13.
dy y x + ( 1 − x²) ¹ ( 39 dx ++ x²)2 (1 (1  x²)
Ex. 14.
( 1  x²)
Ex. 15.
xy² (3y dx + x dy) — (2 y dx − x dy) = 0.
Ex. 16.
( 1 + x²)
Ex. 17.
(5 xy 3 y³) dx + (3x² — 7 xy²) dy = 0.
Ex. 18.
I dy + y cos xsin 2 x. dx 2
Ex. 19.
(xy² + y) dx − x dy = 0.
§ 19
dy xy = axy . dx
dy +1 = tan 1 x. dx
Ex. 20. * (1  x) y dx − (1 + y) x dy = 0. Ex. 21. Ex. 22. Ex. 23.
3xy dx + (x + x³y²) dy = 0. (x² +y²) (x dx + y dy) = (x² + y² + x) (x dy− y dx). (2x + 3y  1) dx + (2x + 3y  5) dy = 0 .
Ex. 24.
(y³
Ex. 25.
(2x²  y) dx + (2 x2y  x)dyo.
2x²y) dx + ( 2 xy² − x³) dy = 0 .
Ex. 26.
Ex. 27.
(x² + y²) (x dx + y dy) + (1 + x² + y²)¹ ( y dx − x dy) = 0 . x x = 0. ( 1 + c³) dx + ev ( 1 − +) dy =
Ex. 28.
xdy + (y  y² log x) dx = 0.
Ex. 29. (x³y + x²y³ + xy² + y) dx + (xªy³ — x³y² − x²y + x) dy = 0 . Ex. 30.
(2√xy — x) dy + y dx:= 0.
CHAPTER III APPLICATIONS 20. Differential Equation of a Family of Curves. Differential equations arise in certain problems in Geometry and the physical sciences. For example, if we let x and y be the rectangular coördinates of a point in the plane, any relation among these, say 4 (x, y) = o , represents some curve, and the value of dy at any point of this curve is dx the slope of the tangent at that point. Starting with a relation that involves an arbitrary constant rationally
$(x, y, c) = 0,
(1)
we have a family of curves , one curve corresponding to each value of c. The differential equation corresponding to ( 1) *, say
dy dx) = 0, 1(x, y, 2
(2)
dy dx corresponding to any pair of values of x and y, we see that ( 2) defines is of the first order.
Since we can obtain from it the value of
the slope of the tangent of that curve of the family ( 1 ) which passes through any chosen point (x, y) .
In case ( 1 ) is of the second degree
in c, we have, excluding exceptional cases, two curves of the family passing through any point, for to each pair of values of x and y correspond two values of c which are distinct, in general .
If, now,
we turn our attention to the differential equation ( 2) , it must give us dy two values of for each pair of values of x and y, in general, since dx * The curves defined by ( 1) are spoken of as the integral curves of (2) . 31
DIFFERENTIAL EQUATIONS
32
$ 20
two distinct curves pass through this point, and, excepting in the points where the curves are in contact, their tangents will be distinct. If, on the other hand, we start with the differential equation and dy it is clear that since at suppose that it is of the second degree in dx' each point of general position, * we have two values for the slope, i.e.
two tangents to the integral curves, we must have, in general, two integral curves passing through each point. Hence it follows that the general integral involves the constant of integration to the second degree.
(See footnote to Ex . 3, § 24. )
Perfectly generally, we can prove, by entirely analogous reasoning, that the integral of a differential equation of the first order and nth degree involves the constant of integration to the nth degree. Ex. 1. Find the differential equation of all circles through the origin with their centres on the axis of x. The equation of this family of circles is evidently x² + y² 2 ax = 0. Here a enters to the first degree. Differentiating, we have x + y
dy — α = 0. dx
Eliminating a, we get
2 xy
dy dx + x²  y² = 0, which is of the first degree.
[As an exercise, the student should integrate this. ]
Ex. 2.
Find the differential equation of all circles of fixed radius
r and with their centers on the axis of x. The equation of this family of circles is (x  a)² + y² = r².
Here
a enters to the second degree. − a) + y. Differentiating, we have, (x —
dy = 0. dx
* By point ofgeneral position , we mean a point at which there is nothing peculiar about the family of curves (such as having two curves tangent to each other there) , or about any of the curves of the family (such as a double point).
$ 20
APPLICATIONS
33
Eliminating a, we get 2 dy +22, which is of the second degree. プ (dx) + Ex. 3. Find the differential equation of the system of confocal conics whose axes coincide with the axes of coördinates. 2 x2 Hint. ― Since the distance of the focus of the conic 2 士 = I a² b2 from the center is √a² = b² it is quite clear that all the conics x2 12 + = I have the same foci, no matter what may be. Hence C this is the equation of a system of confocal conics whose foci are
at the points ( ± √ , o) . eliminated .
Here c is the arbitrary constant to be
Ex. 4. Find the differential equation of the system of parabolas whose foci are at the origin of coördinates and whose axes coincide with the axis of x. Ex. 5. Find the differential equation of the family of straight lines tangent to the circle x² + y² = r². Ex. 6. Find the differential equation of the family of straight lines the sum of whose intercepts on the axes is a constant. Ex. 7. Find the differential equation of the family of nodal cubics (y− a)² = 2 x(x − 1)², • each curve of the family being tangent to the axis of y, and having its node at the point ( 1 , a) . Ex. 8. Find the differential equation of the family of nodal cubics y² = 2 x (x − a) ², each curve of the family being tangent to the axis of y at the origin , and having its node at the point (a, o) . Remark. If the equation of the family of curves involves more than one arbitrary constant, the corresponding differential equation will, of course, be of higher order than the first ( § 3) .
DIFFERENTIAL EQUATIONS
34
§ 2!
Ex. 9. Find the differential equation of all circles tangent to the axis ofy. Ex. 10. Find the differential equation of all central conics whose axes coincide with the axes of coördinates.
Ex. 11. Find the differential equation of all parabolas whose axes are parallel (a) to the axis of x, ( b) to the axis of y. Ex. 12. Find the differential equation of all circles of the same radius r.
21. Geometrical Problems involving the Solution of Differential Equations . Differential equations of the first order arise and must be solved in geometrical problems where the curve is given by properties whose analytic expression involves the derivative of one of the coördinates of a point on the curve with respect to the other. example or two will illustrate : Ex. 1.
An
Find the most general kind of curve such that the tangent
at any point of it and the line joining that point with the origin (which we shall call the radius vector to that point) make an isosceles triangle with the axis of x, the latter forming the base. The tangent of the angle between the tangent line and the axis dy of x is " while that of the angle between the radius vector and the dx
axis of x is y . x Hence we must have
dy = dx
x or x dy +y dx = 0.
Integrating, we have xy = constant, which is evidently an equilateral hyperbola.
APPLICATIONS
§ 21
Ex . 2.
35
Find the most general kind of curve such that the normal
at any point of it' coincides, in direction , with the radius vector to that point.
dx we have Since the slope of the normal is dy
dx dy
y or x dx + y dy = 0. " x
Integrating, we have
x² +y² == c, a constant ; this is evidently a circle. Since the differential equation is of the first degree, a single value of the constant corresponds to a pair of values of x and y. Geometrically, this means that through each point passes one curve of That is, through the family. Thus, if x = 1 , y = 2, then c = 5. the point ( 1 , 2 ) passes the circle x² + y² = 5 , and this is the only circle of the family which does. From the above simple examples the general method of procedure may be seen. It consists, first, in expressing analytically the given property of the curve (this gives rise to a differential equation ) ; secondly, we must solve this equation ; and finally, we must interpret geometrically the result obtained . *
In the Differential Calculus those properties of curves involving differential expressions are usually studied, and their knowledge will be presupposed here . For purpose of convenient reference, the following list will be given : 1° Rectangular coördinates
dy (a) dx is the slope of the tangent of the curve at the point (x, y) ; * The general solution of the differential equation , involving an arbitrary constant, represents an infinity of curves . If we know a point through which the curve must pass, or if in any other way the conditions of the problem determine the constant of integration, either uniquely or ambiguously, one or several curves of the family alone fulfill the requirements.
AL
DIFFERENTI
36
EQUATIONS
$ 21
dx is the slope of the normal at (x, y) ;
(b)
dy = αν (c) Y  y = dx (X− x) is the equation of the tangent at the point (x, y), X and Y being the coördinates of any point on the line ; dx (X x) is the equation of the normal at (x, y) ; (d) Y y = dy dx an d y x dy are the interce pts of the tangent on the (e) x ydy dx axes ; dx dy are the intercepts of the normal on the (f) x + y dx and y + x dy axes ; 2 dx dy are the lengths of the tangent (8) y √1 + ( dt) and x √ √ 1 + (14 dx)
from the point of contact to the x and y axes respectively ; 2 2 dy and x I+ dx are the lengths of the normal (h) yV I+ dx dy from the point on the curve to the x and y axes respectively ; dx is the length of the subtangent ; (i) y dy dy ( j) y dx is the length of the subnormal ;
dy (k) ds = √dx² + dy² = dx◄ I + (2x dx)² = dy ment of length of arc ;
dx is the ele
I+
dy
(1) y dx or x dy is the element of area. 2° Polar coördinates
do
(m) tan
=p
" where
is the angle between the radius vector
dp and the part of the tangent to the curve drawn towards the initial line ; (n) T = 0+ , where the initial line ;
is the angle which the tangent makes with
APPLICATIONS
$ 21
(0) p tan
37
= ρ de is the length of the polar subtangent ; dp
= dp is the length of the polar subnormal ; de 2 do\2 = αρ + p² is the = do dp√1 + p² = dp1 (9) ds = √dp² + p²d0² = do dp element of length of arc ;
(P) p cot
(r) 1 p² do is the element of area ; 2 do is the length of the perpendicular from (s) p = p sin = p² ds the pole to the tangent.
2 I I = I (dp + 4 do (1) 2 Ex. 3. Determine the curves such that the normal (from the point on the curve to the axis of x) varies as the square of the ordinate. In particular find that curve which cuts the axis of y at right angles. Using (h) , we have the differential equation of the curve
税) +(
2 dy = dx. k2 y2 , or = ky², or 1 + dy (dx )² = ky², √k²y²  I
Integrating, we get I logky + √k²y² —
or
= x + c,
I y = 21 (cez + e ), a family of catenaries . 2k C To find the curve of the family which cuts the axis of y at right
angles , we must find that value of c for which do for x = 0 . dx I dy Now CThis equals zero for c = ± 1. Hence the dxx=0 2 I equation of the required curve is ± y = 1½ (ekx (e** + ek²) = ½k cosh ka
==
IAL
RENT
E DIFF
38
S
TION
EQUA
§ 22
Ex. 4. Determine the curve such that the area included between an arc of it, a fixed ordinate, a variable ordinate, and the axis of x is proportional to the corresponding arc.
Using (k) and (1) , we have
I+
2 @ \2 = 'dy = = ky, or 1 + = k²y². dx dx
This is the same differential equation that arose in Ex. 3.
Hence
the catenary has also this property.
Ex. 5. Find the curves such that the polar subtangent is proportional to the radius vector. kdp = do. Using (0) , we have pade = kp, or dp ρ
Integrating, we have
pce , a family of spirals.
Ex. 6. Determine the curves whose subnormals are constant. Ex. 7. Determine the curves whose subtangent at each point equals the square of the abscissa at that point. Ex. 8. Determine the curves such that the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact. Ex. 9. Determine the curves such that the angle between the radius vector and the tangent is onehalf the vectorial angle. Ex. 10. Determine the curves whose polar subtangent is four times the polar subnormal. 22. Orthogonal Trajectories.  A curve, which cuts every member of a family of curves according to some law, is called a trajectory of the family.
Thus, it may cut every curve at a constant angle ; if,
in particular, that angle is a right angle, the trajectory is said to be orthogonal. It is at times possible to find a second family such that
APPLICATIONS
§ 22
39
each curve of the one family is cut at right angles by every curve of the other. If such a pair of families of curves exists, each is said to be a set oforthogonal trajectories of the other. Starting with the first family whose equation is
(1)
(x, y, c) = 0,
we find the corresponding differential equation
dy dx, x, y) = 0, ƒ(dv
(2)
dy which, as we have noted before, defines dx' the slope of the tangent at (x, y) to the curve of the family through that point.
Obviously
dx (3) ƒ(− day, x, y)= 0 will have for integral curves a family such that the slope of the tangent to the curve through the point (x, y) at this point will be the negative reciprocal of that in the case of the corresponding curve of (1) ; i.e. wherever a curve of the one family cuts one of the other their tangents are at right angles, and therefore the curves are said to be at right angles themselves. The integral of (3 ) will then be the equation of the desired family . Ex. 1. Find the orthogonal trajectories of a family of concentric circles. Taking the common center as the origin , the equation of the circles dy = 0. = is x² + y²², and their differential equation is x +y dx Hence the differential equation of their orthogonal trajectories is
dx
dx = O, or
x y dy
x
dy = 0. y
Integrating, we have y = cx, a family of straight lines through the center of the circles.
AL
ENTI
ER DIFF
40 Ex. 2.
ONS
ATI
EQU
§ 22
Find the orthogonal trajectories of the family of circles
through the origin , with their centers on the axis of x. The equation of this family of circles is x² + y² . cx = 0.
We
have seen (Ex. 1, § 20) that the corresponding differential equation is 2 xy dy + x²  y² = o . Therefore, the differential equation of dx dx 2 the orthogonal trajectories is 2 x xy − x² + y² = 0. dy As this equation is obviously derivable from the other by interchanging x and y, its integral must be x² + y²  cy = o, the family of circles through the origin with their centers on the axis of y. [ Let the student verify this result by actually integrating the equation. ]
Ex. 3.
Show that a family of confocal central conics is self
orthogonal. The equation of such a family of conics with their axes taken 1.2 = I, and its differfor axes of coördinates is ( Ex. 3, § 20). + C C λ 2 ential equation is xy((2x) ² + ( x²  y²  x) dy 1/24 — xy = 0. dx dx dx dy we see Since this is left unaltered, when we replace by dx dy
that the family of curves is self orthogonal .
As a matter of fact,
it is well known that a system of confocal central conics is made up of ellipses and hyperbolas, such that through any point there pass one ellipse and one hyperbola, and these cut each other at right angles.
Ex . 4.
(See Ex. 17.)
Prove that the family of parabolas having a common focus and a common axis is selforthogonal.
APPLICATIONS
§ 22
Ex. 5.
4I
Prove that the differential equation of the family of trajec
off x, y) y) tories which cut the integral curves ofƒ y ,, x, (dx dy dx
o at an angle a is
tan a
" x, y dy I + tan α dx
= 0.*
In particular show that the trajectories which cut the lines y = cx at a constant angle a are the logarithmic spirals I I —— tan1 y 2 log (x² + 2) + k = m 0 r = cem.
or
Ex. 6.
Find the trajectories which cut at a constant angle a the
circles through the origin with their centers on the axis of x. Ex. 7.
Find the trajectories which cut at a constant angle a
(other than a right angle) a system of concentric circles. Ex. 8.
Find the
orthogonal
trajectories
of the
parabolas
y² = 4 cx. Find the orthogonal trajectories of the hyperbolas x²y² =c. Ex. 10.
Find the orthogonal trajectories of the similar central
conics ax² + by² = c, where a and b are fixed constants and c the arbitrary constant. In polar coördinates the equation of a family of curves will be
$(p, 0, c) = 0 ;
(1')
and the corresponding differential equation will be
de
p, Ꮎ = 0,
(2') do which defines
at each point (p, 6). dp
* In practice it will be simpler to replace tan a by a single letter, say m.
ENTIAL EQUATIONS
DIFFER
42
We have noted, § 21 (m ) , that, if
§ 22
is the angle between the radius
vector to the point (p , 0) and the tangent at that point drawn in a do · definite direction , tan = ρ If now, is the angle for the curve dp cutting this one at right angles at the point (p, 0) , we have π Hence 4 ' — 4 = ± 7 , or 4' = 4 ± 2
tan
= ' === coty =
I tan
Using primed letters for the second curve , we have then
do' dp'
1 dp' . I do whence do = άρ ραθ' Pp' de'
At the point of intersection of the two curves p' = p, 0' = 0. Hence I do (3') 6, p, 0)= 0 / = 1 1 de 信
is the differential equation of the orthogonal trajectories of (1 ') , de I do since the value of ρ given by it equals that of given by (2 ') . P de dp Note.  Frequently is taken as the independent variable. In this case (2 ' ) will be in the form f
dp P, 0 = 0, and (3 ) will do'
then be
P 0) = 0. 1 pdp, Ex. 11. Find the orthogonal trajectories of the family of lemniscates p² = C COS 2 0. Differentiating and eliminating c, we find the differential equation I of the family to be 1 dp —— tan 20. Hence the differential equaρ do
tion of the orthogonal trajectories is
§ 23
APPLICATIONS de P Pdp
43
tan 20,
de = cot 2 0 do. Ρ
or
Integrating, we find p2k sin 2 0, a second family of lemniscates whose axis makes an angle of 45° with that of the first family. Ex. 12. Find the orthogonal trajectories of the family of cardioids p = c ( 1 — cos 0). Ex. 13. Find the orthogonal trajectories of the family of logarithmic spirals p = e©0. Ex. 14. Find the orthogonal trajectories of the family of curves pm sin me = cm. Ex. 15. Find the orthogonal trajectories of the family of curves I = sin20 + c. Ρ Ex. 16. Find the orthogonal trajectories of the family of confocal 2C and coaxial parabolas p = I  COS Ex. 17. Find the orthogonal trajectories of the family of confocal 22  λ2 conics ρ = c being the parameter, λ a fixed constant. c  λ cos 0' 23. Physical Problems giving Rise to Differential Equations . Problems frequently arise in Mechanics, Electricity, and
other
branches of Physics whose solution involves the solution of differential equations.
As a knowledge of these subjects is necessary to
understand properly the problems that arise in them, we shall restrict ourselves, as far as possible, to problems involving only very elementary principles.
As in the case of geometrical problems, the mode
of procedure is, first the analytic expression of the given data of the
DIFFERENTIAL EQUATIONS
44
$ 23
problem, (this gives rise to the differential equation) ; then comes the problem of solving this equation, with an interpretation of the result.
Frequently there is the additional step of fixing the value of
the constant of integration so as to satisfy the requirements of the problem. The following examples will illustrate : Ex. 1. A body falls vertically, acted upon by gravity only. If it has an initial velocity vo, what will be its velocity at any given instant, and what will be the distance covered in any given period of time ? The motion being rectilinear, a single coördinate, x, will be sufficient to determine the position of the body.
Suppose the position
of the body at the time to to be x (this is called its initial position). dx The velocity, which we shall represent by 7, is and the acceleradt' dv tion, represented byf, is In case gravity acts alone, the acceleradt tion is constant, and this constant is usually designated by g. We dv have now g. dt
Integrating this equation we get v = gt + c.
When to , we have given v = V%• ... c = %, and the answer to our first question is given by v = gt + vo. To find the position of the body at any instant we must integrate
dx =gt + vo• dt The general solution of this is x = 1 gt² + vot + c.
APPLICATIONS
$ 23
45
Since xx when t = 0 , we must have c = xo.
instant t,
Hence at any
1/2 gt² + vot + xo, x =蛋
and the distance covered in the period tis x− x0 = 18t² + vot. / Ex. 2. A particle descends a smooth plane making an angle « with the horizontal plane. The only force acting is gravity. If the particle starts from rest, find the velocity at any moment t, and the distance traveled in the time t [ Hint.  Here the acceleration is g sin a, the component of g in the direction of the motion. ]
Prove
that if a particle starts at rest from the highest point of a vertical circle, it will reach any other point of the circle when moving along the chord to that point in the same time it would take to drop to the lowest point of the circle. TAIT AND STEELE, Dynamics of a Particle. Ex. 3. A particle falls through a resisting medium (such as air) in which the resistance is proportional to the square of the velocity ; what is its motion ?
The equation of motion is then dv g kv². dt Here the variables are separable, and we have dv = = dt. 8 kv²
Putting gk = r², this becomes gdv = dt, or
dv
g + ro dv
dv +
Integrating
•+ m ro So (57
g rv
+ g
dv = 2 dt. rv
= 2 2 S'dt , we have * 0
* A knowledge of hyperbolic functions enables one to effect the integration much more expeditiously, especially if v = 0. ช gdv _ £ tanh rt 2= dx = ~I tanh1 rv dt r So0 2 r272 r g log cosh rt. Integrating again, we have x хо = g r2
ENTIAL
DIFFER
46
ONS
EQUATI
§ 23
[log (g +rv) — log (g + rv ) — log (g − rv) + log (g — rvo)] = 2 t, I log g rv g + ro = 2 t. r \g + rvo g. rv
or
g + rv = _ g + rvo 2rt. 8  ro g rvo Temporarily put the constant
g + rvo.= C. g rvo
Then
Ce2rt  I V= g . r ceart + I dx the position at any time is given by dt' 20 ce2rt  I dx = x xo  g 8S +I dt, where c must finally be replaced .0. cere r S
Since v =
by its value in terms of 7%.
If the body falls from rest, 7% = 0, .' . c = 1 .
[As an exercise, the student may carry out the integration indicated above .] Ex. 4. The acceleration of a particle moving in a straight line is proportional to the cube of the velocity and in the opposite direction from the latter. Find the distance passed over in the time t, the initial velocity being 7%, and the distance being measured from the initial position of the particle, i.e. x。 = 0.
dv = dt
•
kv³, or
dv k dt. 7,3
dx I  I Vo = = 2kt, orv = ; 22 Vo2 dt √2kvot + 1
√2 kv²t +1 and
I
x= kvo
Ex. 5. Find the distance passed over in the time t, if the acceleration is proportional to the velocity.
APPLICATIO
§ 23
NS
47
Ex. 6. One of the important equations in the theory of electricity is
L
di + Ri = E, dt
where i is the current, Z the coefficient of selfinduction (a constant) , R the resistance (a constant) , and E the electromotive force, which This
may be a constant (including zero) or a function of the time.
equation is linear ( § 13 ) .
Find i if (a) E = 0 , ( b ) E = constant,
(c) E = E, sin wt, (E, and w being constants), (d) E = any function of the time, say E (t). Case (c) plays such an important rôle in the Theory of Electricity that its solution is given here in detail.
This equation arises in the
case of alternating currents, where the electromotive force is a peri2π odic function of the time, the period being and the maximum ω value of the electromotive force is E
di dt
R i = Eo sin wt. L L
R An integrating factor is eL'. Ꭱ Rt sin wt dt + c. .. ieL = Eo L ESer
sin wt dt =
Since Seat
a sin wt  w COS wł e at, a² + w²
R sin wt w COS wt R Rt ieL = E, L eL + c L R2 + w² L2
R ωIcost eL + c os uf) cl R² + w²L2
Rsin wt
= Eo
IAL
ENT
ER IFF
48
NS
D
Eo VR2 + w²L2
=
TIO
A EQU
823 wL cos wt
R sin wt
t ·eL + C
√R² + w²L²
Rt Eo eL
sin (wt
) + c,
VR² + w²L2 where
sin
=
..i =
R ωΙ cos & = 2 √R² + w² L2 √R² + w² L² Eo
sin (wt — $) + ce √R² + w²L2
Rt The term ce I usually becomes negligible after a very short interval of time. The current then becomes periodic with the same frequency as the electromotive force. But the two are not in the same phase, the current lagging behind by the angle ø.
CHAPTER IV DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE THAN THE FIRST 24. Equations Solvable for p..
For the sake of simplicity we shall dy adopt the generally accepted notation of replacing byp. Our dx equation may then be written f(p, x, y) = 0.
(1)
If this equation is of the nth degree, we may look upon it as an algebraic equation of the nth degree in p.
Let its roots befi(x, y) ,
ƒ2(X, Y), •••,ƒn(x, y) , then the equation may also be written (2)
[p −fi(x, y) ] [p− ƒ2(x, y) ] ··· [p —ƒ„ (x, y) ] = 0.
Consider now the differential equations arising on equating each of these factors separately to zero. If we can integrate each of these by some method of Chapter II, we can readily get the general solution of ( 1 ) .
Let the solutions of the separate equations arising from (2) be ₁ (x, y, c) = 0 , $₂ (x, y, c) = 0 , ….. , $,n (x, y, c) = o. It is quite clear that
(3)
p1 (x, y, c) p2 (x, y, c) P3 (x, y, c)
is the solution of ( 1 ) .
... Pn (x, y, c) = 0 *
For, the vanishing of the lefthand member of
(3) means the vanishing of one of its factors .
This will cause one of the factors of ( 2 ) to vanish , when substituted in it, and consequently,
the original equation will be satisfied .
Besides, a constant of inte
* If one of the 's is not rational, there will be found certain other irrational ones which , with it, form a set of conjugate irrational functions whose product is rational (see Ex. 2 below) . The result of rationalizing any of the set is the same as this rational product. In practice it is frequently desirable to make use of this fact. The student should verify this fact in the case of Ex. 2. 49
DIFFERENTIAL EQUATIONS
50
gration is involved .*
We see that the constant enters to the same
degree in (3) that p does in ( 1 ).
Ex. 1.
§ 24
(Theorem , § 20.)
² + (x + y) p + xy = 0 ,
or
(p + x) (p + y) = 0. Integrating
dy + x = 0, we get 2 y + x² = c ; dx
and integrating dy + yo, we get log y + x = k, or y = ce*. dx Hence the general solution is ( 2y + x² − c) ( y − ce¯*) = 0.
Ex. 2.
xp²  2 yp — x = 0.
Solving this for p, we get p =
y ± √x² + y² . x
We have, then, to
consider the two equations xp  y = √x² + y² and
xp− y = −√x² + y².
x dy  y dx = √x² + y²
dx may be written
x dy  y dx = dx which gives, 2 x x2 VI + √₁ + (1)²
on integrating, 2 y y = log x + k, or x + log [²{ + √ + + ( 1) ]³ = >
2 I+ y
 cx = 0.
* Since (3) is a solution by having each of its factors separately satisfying the differential equation ( 1) , it may be asked why we use the same constant c in all the factors. If we look upon equation ( 1 ) as equivalent to the n separate equations arising on equating each of the factors of (2) to zero (which, in fact, it is) , then we really have n equations to solve, and the various factors of (3 ) , involving distinct constants, are the solutions of these. But if we require the general solution of ( 1 ) to be given as a single expression, first, there is no room for more than one constant of integration (§ 4) , and then, there is no loss in using the single constant throughout, from the very fact that ( 3 ) is a solution by virtue of each factor separately satisfying the equation.
THE FIRST ORDER AND HIGHER DEGREE
§ 24
51
Integrating xp y =  √x² +y², we get in a like manner I
log
= log
+ VI + x
y
+
+
2
y x
= = log
It
y x
I
 log x + k, or y x
y 2 I+   cx = 0. x
Hence the solution is y2 CX cx 0, x) x)* — « ] [ + √¹ + ( } − √¹ + ( ˆ˜  « ] = • or
c²x²  2 cy — I = 0. Ex. 3. y² + p² = 1. Here p
± √12 .
Solution of
dy = dx is sin¹y = x + c , VI y² dy
= dx is y = cos (x + c).
or y = sin (x + c) ; while that of VI  y² Since
is perfectly arbitrary, either one of these is sufficient as the
general solution of the equation ; * for sin (x (x+ + cc +· + 21 ) = cos (x + c) . * Since centers in a transcendental way, and not to the second degree in the solution y sin (x + c) , or y = cos (x + c) , we seem to have an exception to the rule of § 20. It is really not such. That rule presupposes that the constant of integration enters algebraically. We can make our solution conform to the rule by writing it in the form (sinly  x  c) (cos  ly  x  c) = o. As a matter of fact the rule was based on the fact that through each point of general position pass two integral curves of a differential equation of the second degree. So here, of the infinite number of values of c that satisfy y = sin (x + c) for a given pair of values of x and y, only twc will determine distinct curves. A more elegant form than the one above, in which c enters algebraically and to the second degree, may be gotten thus : Since c is any 2k I  k2 ; then cos c = number, we may write sin c = ; whence, remembering I + k2 1+ k2 that sin (x + c) = cos c sin x + sin c cos x, the solution takes the form y— cos x − 2ksin x + k2 ( y + cosx) = 0.
DIFFERENTIAL EQUATIONS
52
§ 25
Ex. 4. (2 xp − y)² = 8 x³.
Ex. 5. ( 1 + x²) p² = 1 . Ex. 6. p³
(2x +y²)p² + (x² −y² + 2 xy²) p − (x² − y) ² = 0.
25. Equations Solvable for y. 
If the equation can be solved for
y, the following method will frequently be found useful. y, we have
Solving for
y = 4 (x, p).
(4) Differentiating this we get
ay ay dp pax + o d ' p x
(5)
a differential equation which is really of the second order, but since y is no longer present, it may be looked upon as an equation of the first order in the variables x and p. It may happen that we can integrate this equation. Suppose its solution to be w (x, p) = c.
(6) Eliminating (7)
from (4) and (6) we have (x, y, c) = 0,
which is a solution of (4) ; and , since it involves an arbitrary constant, it is the general solution. We know that (7 ) is the solution of (4) from the following considerations : Since (5) is the derivative of (4 ) , every solution of (4) is a solution of (5) , considered as an equation of the second order in x and y.
Since (6) is a solution of (5) , every solution of (6) looked
upon as a differential equation in x and y is a solution of (5) . (4) and (6) are known as first integrals of (5) . Since (4) contains y and (6) does not, these two first integrals are evidently independent. Equation (4) has an infinite number of integral curves, * and so has * We use this geometrical mode of expression, not because it is essential to the argument, but because it is simpler.
825
THE FIRST ORDER AND HIGHER DEGREE
(6) for each value of c .
53
To find the curve or curves common to the
two families of integral curves we shall have to find the equation of the locus of those points for which (4 ) and (6) determine the same * value ofp. But this locus is evidently gotten by eliminating from (4) and (6) .
For each value of c we get thus one integral curve of is an arbitrary constant, we have the general
(4). Hence, when solution.†
Note. This method applies equally well to equations of the first degree. See Ex . 4. Remark.  At times it is easy to integrate (6) . Doing this, a relation • (x, y, c, c') = 0, (8) involving two constants, results. This is the general solution of (5 ) and must therefore contain that of ( 4 ) for some relation between c and c'. This relation can be found by substituting ( 8 ) in (4) and noting what condition is imposed, so that the equation be satisfied . In actual practice, however, this method will generally not be as desirable as the one given above. Ex. 1. 2px  y + logo ; or (1) y = 2px + log þ. Differentiating, we get dp p= 2p +(2x + 1) dx' 1 / 2x+ Ι p dx + 2 x dp + = dp = = 0. P
or
An integrating factor is seen, by inspection, to be p. we have p² dx + 2px dp + dp = 0.
Using this,
Integrating we have (2) Eliminating
p²x +p = c. between ( 1 ) and (2 ) , we have the required result.
* As already mentioned (§ 20) , a differential equation of the first order may be looked upon as defining p, the slope of the integral curve, at each point (x, y) . + Since the process of eliminating from (4) and (6) may introduce extraneous factors, and errors may enter in other ways , it is desirable to test the result (7) , by finding out whether it actually satisfies the equation (4) .
54
DIFFERENTIAL EQUATIONS
825
Remark. In this case, while it is perfectly possible to perform this elimination [ since ( 2 ) can be readily solved for p, and the latter value can be put in ( 1 ) ] , the result will not be very attractive in form . It is simpler to say that (1 ) and ( 2) taken together constitute the solution, in that, from them, we can express x and y in terms of p, which may be looked upon as a parameter. Thus from (2) we have x = Cp2 2 p) + log p. and ... y = ² (c( c− p) P Such parametric representation is frequently resorted to ; thus, for example, the parametric equations of the ellipse
x = a cos 0, y = b sin 0, where is the eccentric angle, and also the usually adopted equations of the cycloid sin 0) , x = a (0 = a (1 — cos 0) . J
Ex. 2. 4 xp2 + 2 xp  y = 0 ; or
y = 2 xp + 4xp². Differentiating, we have on collecting terms, (4p + 1 ) (2x dx dp +p) = 0.
Neglecting the factor 4 + 1 , whose significance we shall see later (§ 32 , note), we have, integrating the other factor, xp² = k². Hence the solution is y = 2k√x + 4k², or putting 2k = c,
y = c√x + c². Rationalizing this, we have (y² )² = c²x, or putting
(y − C)² = Cx.
= C,
§ 26
THE FIRST ORDER AND HIGHER DEGREE
= 0. Ex. 3. xp²  2 yp  xo
55
(Ex. 2 , § 24.)
Ex. 4. p + 2xy = x² + y².
Ex. 5. y = xp + x¹p².
Ex. 6.
2 + 2 xp  yo.
26. Equations Solvable for x.A method, entirely analogous to that of the previous paragraph, can be deduced in case the equation can be solved for x. Suppose the equation in the form
x = 0 (y, p).
*(9)
Differentiating with respect to y we get dx =
(10)
dy
I  20 ao dp . + ду ap dy
p
Here x no longer appears, and we may look upon this equation as one of the first order in y and p. If we can integrate this, we obtain a relation involving an arbitrary constant,
(11) and on eliminating solution .
x (y, p, c) = 0 ; between (9 ) and ( 11 ) we have the general
Ex. 1. x +py (2 p² + 3) = 0. Ex. 2. a²yp²  2 xp + y = 0 . Ex. 3. xp22 yp  x = 0.
(Ex. 3 , § 25.)
Ex. 4. p³  4 xyp + 8 y² = 0 . Ex. 5. Find the family of curves for which the length of the normal (from the curve to the axis of x) is equal to the square root of the length of its intercept on the axis of x.
56
DIFFERENTIAL EQUATIONS
827
27. Clairaut's Equation .* — If the equation is of the form (1)
y = px + ƒ(p) ,
where f( ) is any function ofp, the solution is gotten so readily that especial attention should be given to this form of the equation, in order that it may be recognized at once.† Using the method of § 25 , we have
dp p = p + [x +ƒ' (p ) ] dx' dp df dx [x +ƒ'(p) ] = 0 .
or
Neglecting the factor x +f ' (p) , which involves no differential expressions (see § 32 , note) , we have
dp = o, whence p = c. dx Putting this value in (1 ) we have (2)
y = cx +f(c),
which is the general solution of ( 1 ) .
 y)² Ex. 1. (px − = p² + 1 . Solving for y,
y = px ± √p² + 1 .
This being in Clairaut's form, its solution is known at once to be
y = cx ± √c² + 1 , or
(cx − y) ² = c² + I.
* This form of equation is named after Alexis Claude Clairaut ( 1713–1765) . He was the first to apply the process of differentiation ( §§ 25 , 26) to the solution of equations. His application of this method to the equation that bears his name was published, Histoire de l'Académie des Sciences de Paris, 1734. † The student should be able to recognize this equation , not only when it is solved for y, as it is in ( 1) . Obviously, what characterizes this form of the equation is that x and y occur only in the combination y px. Hence any function of y  px and pequated to zero, say f(y  px, p) = o is a Clairaut equation, and its solution is f(y cx, c) = o. (See Ex. 1.)
§ 27
THE FIRST ORDER AND HIGHER DEGREE
57
Ex. 2. 4 e²² + 2 xp − 1 = 0. Put et, then
dy = I dt and the equation becomes p:= dx 2 t dx
t= x
dt (dt)?2 + dx dx
its solution is t = cx + c², or e²y := cx + c². Remark.  At times, as in the case above , a transformation can be found to simplify very materially an equation which will not yield directly to any of the previous methods . formations are not always obvious.
Unfortunately these trans
Experience, and frequently that
alone, will help one in making a proper choice. Ex. 3. 4 e²² + 2 e²xp − e²x = 0 .
Put ex = u, e²v = v, then p 
v=u
whence its solution is
u dv " and the equation becomes 2 v du
dv\2 dv + " du du
e2y = ce² + c².
Note. One's first impulse would be to try e2x = u and e²y = v. Our equa2 dv • This is not in Clairaut's form ; but it tion then becomes v = 2 u + 4u du (a ) can be integrated (see Ex. 2, § 25 ) , so that this transformation is also effective. Ex. 4. e²² + (e²x + e³x)p — e³x = 0 . Ex. 5. xy²p² — y³p + x = o .
( Let x² = u, y² = v .)
Ex. 6. (x² + 1²) ( 1 + p)² − 2 (x + y) ( 1 + p) (x + yp) + (x + yp)² = 0. (Let x + y = u, x² +y² = v.)
Ex. 7. y = 2px + y²p³.
(Let y² = v .)
Ex . 8. a³yp²  2 xp + y = 0.
(Ex. 2 , § 26. )
Ex 9. (xp − y)² = x²( 2 xy — x²p) . Let y=vx
(Let 2 x = u, y² = v.)
TIAL
EREN
DIFF
58 28. Summary.
S
TION
EQUA
$ 28
Given a differential equation of the first order
and higher degree than the first, there are three methods * which suggest themselves, to be tried in actual practice in the following order : 1° Solve for p, and then solve the resulting equations of the first degree (§ 24) . 2° Solve for y, differentiate with respect to x, integrate, and between this solution and original equation ( § 25) . eliminate
3° Solve for x, differentiate with respect to y, integrate, and eliminate between this solution and original equation ( § 26). Clairaut's form ( § 27) has been given special prominence in this chapter because of the ease of finding its solution . It is, of course, solved by method 2 °. If none of the above methods work, a substitution must be sought to bring the equation into manageable shape. There are certain cases when we can tell in advance that some or all of these methods work. Here the difficulties are those of Algebra or the Integral Calculus, and not of the Differential Equations. For example, consider the following cases : (a) If the equation in p is algebraic and all the coefficients are homogeneous and of the same degree in x and y, then, on dividing by the leading coefficient, all the coefficients are homogeneous and of zero degree . Hence, if we can solve for p (which is an algebraic process) , we shall find p as homogeneous functions of x and y of degree zero, and the resulting equations, when subjected to the transformation y = vx, will have their variables separated ( § 10) and are solvable by quadratures. Again, since after dividing by one of the coefficients of the equation the y = o, if we solve for y, equation is a function of p and x' say fp, x , ( カ ) ( or 12/2) we get y = x (p) . Differentiating, we have, dx 'dp " or p= 4 ( p) + x¥' (p) dp dx p4(p) x where the variables are separated .
* It is almost needless to remark that these methods are not mutually exclusive Two, or all three, methods may be applicable to some equations.
§ 28
THE FIRST ORDER AND HIGHER DEGREE
59
Hence, in this case methods 1° and 2° both work, provided we can solve for p and for y. (b) If x is absent, so that the equation is of the form ƒ(p, y) = o, solving for p, dy = dx. = (y), or we get p = dy dx 4(1) Again, solving for y, we get y=4 (p) , and differentiating, we have p =
'(p) dx',
or 4' (p) dp = dx, where the variables are separated. Here again methods 1° and 2° both work. (c) If y is absent , equation is f(p, x) = 0. Let the student show, as an exercise, that in this case methods 1 ° and 3° both work, provided we can solve for p and for x. (d) If the equation is of the first degree in x and y, thus, xfi (p) +Yƒ½ (P) +ƒ³ (p) = 0 *, it is readily seen that method 2° works. For, solving for y, we have ' = x¥1 (p) + ¥2 (P) . y
Differentiating, we get
p = 41 (p) + [x¥1' (p) + ¥2′ (p) ] dp dx . Considering
as the independent variable, this may be written, dx + x 41'(1) = 42'(p) dp 41 (p) pp  41 (p)'
which is linear and can be solved by quadratures (§ 13) . Ex. 1. y² (1 +p²) = a². Ex. 2. yp = (x — b)p² + a.
Ex. 3. x² + x²yp + 1 = 0 . Ex. 4. 3x  6 yp + x + 2y = 0.  Ex. 5. y = p²(x + 1) . Ex. 6. (px  y) (py + x) = a²p.
(Let x² = u , y² = v .)
Ex. 7. p² + 2py cot x = y². * Clairaut's equation is a special case of this.
DIFFERENTIAL EQUATIONS
60 Ex.
8.
Ex.
9. x2p² — 2 (xy + 2p) p + y² = 0 .
§ 28
(1 + x²)p²  2 xyp + y² −1 = 0.
Ex . 10. y = xp +
². x2
( Let x² = u, y² = v.)
Ex. 11. x²p²  2 xyp + y² = x²y² + xª.
Ex. 12.
y  xp ; = f(√x² + y²).
(Let x = p cos 0, y = p sin 0. )
Vi + p Ex. 13. Find the equation of the curves for which the distance of the tangent from the origin varies as the distance of the point of contact from the origin.
Ex. 14. Find the equation of the curves such that the square of the length of arc measured from a fixed point is a constant times the ordinate of the point. (Let the constant factor be 4k .) Ex. 15. Find the equation of the curves down each of whose tangents a particle, starting from rest, will slide to the horizontal axis in the same time.
[As in Ex. 2 , § 23, we have that the distance I p covered in the time t equals  gt2 sin a. Here sin α = 9 and 2 Vi + p²
the distance covered is
y √1 +£² , § 21 , (8) .] P
CHAPTER V SINGULAR SOLUTIONS 29. Envelopes . We have noticed before that (x, y, c) = 0, where is an arbitrary constant, represents a family of curves, to each value of c corresponding some definite curve (provided c enters rationally, which we shall suppose to be the case throughout this chapter) .
So that if we pick out some curve corresponding to a
definite value of c, we can suppose our attention directed to the different curves corresponding to c as it varies continuously. * We shall be interested in the locus of the ultimate points of intersection of each curve with its consecutive one. By the ultimate points of intersection of a curve with its consecutive one we mean the limiting positions of the points of intersection of a curve with a neighboring one as the latter approaches coincidence with the former. (Thus in the case of the family of circles referred to in the footnote, the ultimate points of intersection of two consecutive curves are the extremities of the diameter perpendicular to the axis of x.) find the equation of this locus, we proceed as follows : If (1)
To
(x, y, c) = 0
is the equation of a curve corresponding to some chosen value of c,
(2)
(x, y, c + Ac) = 0
* Thus, for example, consider the family of circles of fixed radius whose centers all lie on the axis of x ; their equation is (x − c) ² + y² = r². When co we have the circle whose center is at the origin, and as c increases we get circle after circle whose center is (c, 0). 61
DIFFERENTIAL EQUATIONS
62
§ 29
will be the equation of a neighboring curve, Ac being a finite constant quantity, different from zero. To find the points of intersection of these two curves we have to solve ( 1 ) and ( 2 ) for x and y, or we may replace (2) by a constant times their difference, i.e. by
1 (x, y, c + Ac) — p (x, y, c) = 0. Ac
(3)
To obtain the ultimate points of intersection of the curve with its consecutive one we combine ( 1 ) with what (3) becomes when we let Ac approach the limit o, i.e. with
dd (x, y, c) ac
(4)
If we were to solve ( 1 ) and (4) , we should actually obtain the intersections of the curve with the consecutive one. But what we want
α
ÍI
b
III
FIG. I is the locus of these points, for all values of c. gotten by eliminating c between ( 1 ) and (4) . as the envelope of the family (1).
This is evidently
This locus is known
A property of the envelope which
we shall have occasion to use is : At each point of the envelope there This is immediately obvious
is one curve ofthe family tangent to it.
Suppose ( I) , ( II) , ( III) are three curves of the from the figure. family which ultimately become coincident. a becomes an ultimate point of intersection of ( I) and ( II) , and b of ( II) and (III) .
Hence
they are both points on the envelope, and the line joining them becomes ultimately a tangent to the envelope . But they are also both on the curve (II ) , so that the line joining them also becomes a
$ 30
SINGULAR SOLUTIONS
63
tangent to ( II) . That is to say, ( II) is tangent to the envelope at the limiting position of a and b .* Ex. Find the envelope of the family of circles referred to in the footnote, page 61 . 30. Singular Solutions. — Suppose now that (1)
$ (x, y, c) = 0
is the solution of f( p, x, y) = o.
We have already noted that,
looked at geometrically, this means that the slope of the tangent at a point (x, y) of the curve of the family defined by ( 1 ) passing through that point is exactly the value of p given by ƒ ( p, x, y) = 0 for that pair of values of x and y. But we have just seen that the tangent at any point of the envelope of the family of integral curves coincides with that of the integral curve through that point. It follows, then, that the equation of the envelope will satisfy the differential equation, and is consequently a solution. Moreover, since the envelope is usually not a curve of this family, i.e. its equation cannot be gotten from ( 1 ) by assigning a definite value to the parameter, the equation of this envelope is a solution , distinct from the general It contains no arbitrary constant, and is not a particular solution. It is known as the singular solution. I Ex. 1. y = px + Þ
solution.
This being Clairaut's equation , its solution is I y= cx + =, or
Ex  cy + 1 = 0 .
*This theorem ceases to hold in case the limiting position of a is a singular point on (II), such as a double point or cusp. See Fig. 2, § 33. While, as we shall see (§ 33) , if each curve of the family has a singular point, the locus of these satisfies the geometrical as well as analytic requirements for an envelope, it is usually customary to apply the term envelope to that part of the locus of ultimate points of intersection of the curves of the family which has a curve of the family tangent to it at each point.
64
DIFFE
EQUAT
RENTI
AL
831
IONS
Differentiating with respect to c, we have
2 cx y = 0. Eliminating dz + c ' = — ced=¬* + c ', or
y + cx = c'xe3x3
Ex. 7. x² ( 1 − x²) D²y — x³ Dy  2 y = 0. Ex. 8. x² D³y  5 x D²y + ( 4 x² + 5) Dy — 8 x³y = o.
Ex. 9.
d'y 'dy 2 +f(x) dy + $ ( y) = 0. dx dx2 dx
* This type of equation was first treated by Joseph Liouville ( 18091882) . Let the student show that it is the differential equation corresponding to a primitive of the form F (y) = a Þ (x) + b, where F and ♣ are any functions of their respective variables, and a and b are the arbitrary constants to be eliminated.
HIGHER ORDER THAN THE FIRST
$ 61
By inspection
143
dy 1 is seen to be an integrating factor. dx
Introducing this, we have 'dy 1 d²y dy =: 0 ; dx dx2 +ƒ(x) + $ (1) dx
whence
log
dx +Sƒ(x) dx + S$(y)dy = c, or
eSoudydy = ceS (x)dxdx ;
and Se$$(w)»dvdy = cSe−51(a)dzdx + c'.
Ex. 10.
2 d2y = 0. + 2 cot x dy + 2 tan y dx dx² dx
61. Transformation of Variables . —  In case the equation to be integrated does not come under any of the heads already treated , it is possible, at times, to reduce it to one of them by a transformation . No general rule for this can be formulated . The form of the equation will frequently suggest the transformation to be tried. 2 Ex . 1. x²yd² + (xdy. − y) = 0. dx2
The set of terms
x dy (a dx
y
suggests the transformation y = vx.
Making this transformation, the equation becomes (after dropping the factor ³) 2 dv d2v +20 XV + x dv = 0. e dx dx2 ) (ddx
DIFFERENTIAL EQUATIONS
144
§ 62
This is exact, and has for first integral XV
dv I + = v² = c. dx 2
xv2 XV² = C₁x + €29
This is also exact, giving or
y² = 4₁x² + C₂X. [A less obvious transformation is y² = V .
Let the student solve
the problem by making this transformation . ]
Ex. 2.
dy dx²
Ex . 3. y d'y dx²
2 x dy y = 0. dx 2 dy = y²logy  x²y². dx
Ex. 4. sin2 x d2y 2y = 0. dx2
[Let log y = v, i.e. y = e".]
[ Let cot x = z. ]
If the more obvious transformation sin x = z is made, the resulting equation can be made exact by multiplying by a proper power of %, and can then be integrated.
62. Summary.
The number of classes of differential equations
of higher order than the first for which a general method of solution is known is very small .
We can tell by inspection
1° when the dependent variable is absent ; let the lowest ordered derivative * that appears be a new variable ( § 57 ) ; 2° when the independent variable is absent ; let the first derivative of the dependent variable be a new variable, and consider the * Provided this is not also the highest ordered derivative that appears. If such is the case, let the next lower ordered derivative be a new variable.
§ 62
HIGHER ORDER THAN THE FIRST
145
dependent variable as the independent one ( § 58) ; in particular, if
the equation has the form dy = f(y) , this method leads to the obdx2 dy dx (§ 58, Remark). vious integrating factor dx 3° If the equation is linear, and a particular integral y can be found when the righthand member is made zero, let y = y'₁ , and dv = ($ 59). in the transformed equation put dx 4° If the equation is linear and of the second order, the methods of Chapter VIII may apply ( § 55) . If none of the above cases occur, test the equation for exactness (§ 60) .
Should this not prove to be the case , some special device
must be resorted to, such as finding an integrating factor (§ 60) , or finding some suitable transformation ( § 61 ) . As a final resort, the method of integrating in series may be tried (§ 74) .
Ex. 1.
d²y  'dy 2 + I. dx2 dx
Ex. 2. (1— x²) d2y x dy = 2 dx2 dx
Ex. 3.
d²y + y dy = 0. dx dx2
dy + 6y = 0. Ex. 4. ( 1 + x³) ² + 9x² d²y + 18 x dx2 dx dx3
Ex. 5. (x2  x)
d'y dy + 2y = 0 . + (4x + 2) dx dx2
Ex. 6. y ( 1 — logy)
2 d'y = 0. + (1 + logy) ( dx2 dx
DIFFERENTIAL EQUATIONS
146
$ 62
d2y I dy = 0. + dx2 x dx
Ex.
7.
Ex.
d'y 'dy 2 dy 8. x (x + 2y): + 2x +4 (x + y) dx + 2y + x² = 0. dx2 dx
Ex.
9.
(dy 2 d'y + + I = 0. dx2 dx
) day _ 1 dy + x² = 0. x²) Ex. 10. (1  x² dx² x dx
Ex. 11. 4 2 d³y d'y dy = 0. + 8x + dx2 dx dx3
Ex. 12. sin x
d'y dx2
COS X
dy + 2 sin x ⋅ y = 0. dx
Ex. 13. Determine the curves in which the radius of curvature is equal to the normal, (a) when the two have the same direction, (b) when they have opposite directions .
3 2 2 dy dx ] " The radius of curvature = ±[ · + (** ) ' d'y dx²
The normal
being
supposed drawn toward the axis of x, when it and the radius of d'y have opposite curvature are drawn in the same direction, y and dx2 signs ; and when drawn in opposite directions, y and same sign.
d2y have the dx2
§ 62
HIGHER ORDER THAN THE FIRST
147
Ex. 14. Determine the curves in which the radius of curvature is twice the normal, (a ) when the two have the same direction, (b) when they have opposite directions. Ex. 15. Find the curves whose radius of curvature is k times the cube of the normal. Ex. 16. A particle which sets off from a point of the axis of x, at a distance a from the origin, moves uniformly in a direction parallel to the axis of y.
It is pursued by a particle which sets off at the
same time from the origin, and travels with a velocity which is n times that of the former.
Required the path of the latter.
[ This path is usually referred to as the curve of pursuit.
Its
differential equation may be obtained from the following considerations : Let (x, y) be the coördinates of the pursuing point, ( , of the point pursued . given ( 1 ) ƒ (§, n) := 0.
) those
The path of the latter being known, we have
Since the point pursued is always in the tandy gent to the curve of pursuit, we have (2 ) ŋ — y = dx ( x). (1) and dy If the velocities of the (2 ) determine έ and η in terms of x, y, dx point pursued and pursuing point are as in, we have
n√d§² + dŋ² = √dx² + dy² ; or taking x as the independent variable, 2 2 dn n + = I + dy dx dx dx Substituting in this the values of έ and η ŋ from ( 1 ) and (2 ) , we obtain the differential equation of the curve of pursuit .]
Ex. 17. Find the velocity of the weighted end of a simple pendulum of length 7, swinging in a vacuum, if at the time to , v = 0 and = α, where a is not so small that sin a may be replaced by a as a first approximation .
(See Ex. 17 , § 52.)
NTIAL
148
DIFFERE
NS
EQUATIO
§ 62
Ex. 18. A particle moves in a straight line attracted by a force varying inversely as the square of the distance . [Equation of motion k2 d2x is If it starts with zero velocity at a distance a from x2° dt the center of the force, (a) find its velocity at any point in its path, (b) find the time required to reach that point, (c) how far will it have to move in order to acquire the same velocity with which it would arrive at the point a if it had started to move from infinity with zero initial velocity. (d) Since gravity acts according to the above law, find the velocity with which a body (a meteorite, for example) will strike the surface of the earth if it falls from a distance h above the surface. [Acceleration due to gravity at the earth's surface is usually designated by g.
Hence k² = gR², if R is the radius of the earth .]
CHAPTER X SYSTEMS OF SIMULTANEOUS EQUATIONS 63. General Method of Solution .  It is proved in the general theory of ordinary differential equations that a system of n equations involving n dependent variables can, in general, be solved (§ 70). We shall consider here the case of n = 2 , the method admitting of being extended to any number.
Let the equations be (y)r, 1] = 0,
(1)
fi[(x)m,
(2)
ƒ2[(X)m+p, (Y)s, t ] = 0,
where the highest ordered derivatives of x appearing in ( 1) and (2) are respectively m and m +p , those of y are r and s, and independent variable. Differentiating (1)
Of course
times, we get successively
(3)
fs[ (x) m+ 1, (y)r+1, t] = 0,
(4)
ƒ₁[(x)m+2, (y)r+2, t] = 0,
(
+2)
We now have
is the
may be zero.
ƒp+2 [ (X)m+p, (V)r+p, t] = 0 . + 2 equations from which to eliminate x and all of
its m +p derivatives .
In general (unless m = 0) this will not be
sufficient, for we must have one more equation at our disposal than the number of quantities to be eliminated. 149
We proceed now to
150
DIFFERENTIAL EQUATIONS
differentiate both ( 2) and (p + 2 ) .
§ 64
Since this introduces two new
equations and only one new derivative of x, we see that, by repeating the process the proper number of times, the number of equations will exceed that of the quantities to be eliminated by unity. Performing the elimination, we have a single equation in y.
Integrating this
and substituting the value of y in ( 1 ), we have an equation in x only, which must then be solved . Remark. — It is almost needless to add that we may first eliminate y and its derivatives, and then solve for x. Or, we may solve for x and for y separately. In this case the constants of integration arising are not all independent. The relations among them can be found by substituting in one of the equations ( 1 ) and (2) . 64. Systems of Linear Equations with Constant Coefficients .This method can be carried out very readily in case the equations are linear and the coefficients constants. Thus consider the example
dx ― dy + x = cos t. dt dt
dx  dy + 3x  y = e². dt dt2 These may be written
(1)
(D + 1) x  Dy = cost,
(2)
(D² + 3) x − (D + 1) y = e² .
Differentiating (1 ), we have
(3)
(D² + D) x  D²y = — sin t.
We must eliminate x, Dx, D'x ; this requires four equations. Hence we must differentiate ( 2 ) and (3).
This gives rise to
(4)
(D³ +3 D)x − (D² + D)y = 2 e²,
(5)
(D³ + D²) x  D³y == = cos t.
$ 64
SYSTEMS OF SIMULTANEOUS EQUATIONS
151
We have now five equations from which we can eliminate the four quantities x, Dx, D'x, D³x.
By taking  3 × (1 ) , I × (2) , 1 × (4),
IX (5) , and adding, we get (6)
— D² + D  1) y = 3 e²t (D³ et . 2 cost,
which is a linear equation in y only, and can be solved readily. Before doing so, however, we shall see how (6) can be gotten directly from ( 1 ) and ( 2) .
Since , when looked upon algebraically,
(3 ) and (5 ) are respectively D and D2 times ( 1 ) , and (4) is D times (2 ) (temporarily supposing their righthand members to be zero) , the above method of elimination amounted to subtracting (D² + 3) times (1 ) from (D + 1 ) times (2).
But this is precisely the method
we would have pursued in eliminating x from ( 1 ) and (2 ) had D and its powers been algebraic quantities instead of operators . Now, so long as the equations are linear with constant coefficients, this process is always allowable, since it involves only the operations of addition , subtraction, and multiplication with the operator D. Hence we need only write our equations in the form of ( 1 ) and ( 2 ) , solve them as algebraic equations, remembering, however, that D is an operator in case there are any terms in the righthand members. In practice it is frequently convenient to use determinants. (2 ) for y, we have
D +I

D
Thus solving ( 1 ) and
D +I
COS
y=
D2 + 3
(D + 1 )
D2 + 3
or
(6)
(D³ — D² + D − 1 )y = 3 e² — 2 cos t.
The complementary function is Y= c₁et + c₂ sin t + c3 cos t. For the particular integral, try U= ae² + btsin t + ct cos t. I = 1, c = 1; b= Substituting this for y in ( 6) , we find that a = 3." b 2 2 5 I 32 + I t sint + = tcost. (7) •. y = Get + €2 sin t + c cos t + 2 2 5
DIFFERENTIAL EQUATIONS
152
$ 64
To find x we may substitute this value in either ( 1 ) or ( 2 ) , and solve the resulting equation in x. * Or we can treat x exactly as we did y, that is, solve ( 1 ) and ( 2 ) directly for x. Doing this we have
D +I

COS t
D
x=
D² + 3
 (D + 1)
e2t

D COS t D = D +I e2t − (D + 1 )
or
(8)
(D³ — D² + D  1 ) x
2 e + sint  cost.
Solving this, we have
(9)
x = c ' et + c₂ sin t + cg' cos t + 2e2t + t cost. 2 5
But these constants are not independent of those in ( 7 ) . They • may be found by substituting ( 7 ) and ( 9 ) in either of the original I equations and equating coefficients. Doing this, we find ' = = cv 2 I I I €2' = 12 (€2 − €3) + 3 , ca ' = 12 (€2 + €3) + 1 · * In general, in solving for the variable first eliminated , it is necessary to solve a differential equation. For example, if we put the value of y given by (7) in ( 1) , we have an equation of the first order to solve ; if we put it in (2) , we have an equation of the second order to solve. The new constants of integration that arise now are not arbitrary, but must be determined so that the other equation is also satisfied . This is done by substituting in the other equation , and equating coefficients . In this particular example it would have been simpler to have solved for x first. The value ofy could then be gotten immediately from the equation resulting from subtracting (1 ) from (2) . Let the student do this. We see by this method of solution that the differential equations in x and in y, each resulting from the elimination of the other variable, have the same lefthand members, and that the complementary functions are therefore of the same form in the case ofthe two variables. This is obviously true in the case of n dependent variables defined by n linear equations with constant coefficients.
§ 65
SYSTEMS OF SIMULTANEOUS EQUATIONS
153
Hence the general solution of our system of equations is 4 x = 2 Get + ( 2 Cg − 2 C3 + 3 ) sin t + ( 2 c₂ + 2 3 + 1) Cost
8 +
et + 2 t cost, 5
I y = Get + c₂ sin t+ c cos te² +  t (sin t + cos t). 2 5
3
dx + 3x + 2y = et, dt
Ex. 1. 4x
2 d²y dt2
3
dy + 3y = 3t. dt
dx ― 4y = 2t, dt
Ex. 2. 4
dx dy +2 3x = 0. dt dt
d2x dt2
3x  4y = 0,
Ex. 3.
d2y + x + y = 0. dt2
65. Systems of Equations of the First Order.
 If the equations
are of the first order, we can suppose them solved for the first derivatives of each of the dependent variables . case of two dependent variables .
[ We shall consider the
But the methods here brought out
obviously apply to the case of n such variables. ]
(1)
dx P (x, y, t) = dt R( x, y, t)' dy = Q(x, y, t) R(x, y, t) dt
Let the system be
DIFFERENTIAL EQUATIONS
154
The general method of § 63 applies.
But in certain cases the
solution can be brought about very much more readily. of these cases that we shall consider now. more symmetrical form dx = dy dt (2) P е R
§ 65
It is some
( 1 ) can be written in the
1º One of these equations may involve only two of the variables, or it may be possible, by a proper choice of a pair of members of (2 ) , to strike out a common factor so as to obtain an equation involving only two of the variables. Thus, to fix the ideas, suppose dx  dy that does not appear in or can be removed from it. We P e have, on solving this ,
(3)
(x, y) = c1.
If this can be done a second time, so that a second relation of the form
4 (y, t) = c2
(4)
can be found, then the complete solution consists of the two relations (3) and (4) .
Ex. 1.
dx dy = yt tx
dt
xy
From the first two members we have x²  y² = 61• From the last two members we have y²  t² = C2• } (Using the first and last members we get 2  x² = c.
But this is
obviously not distinct from the other two.) 2° If we can find only one integral expression of the above type, say (3), we can, by means of it, express one of the variables in terms of C1, and the other ; thus, to fix the ideas, we can solve ( 3) for x in dy  dt we have an terms of c₁ and y. Substituting this value of x in Q R'
$ 65
SYSTEMS OF SIMULTANEOUS EQUATIONS
equation involving y, t, and the constant c₁.. second relation
155
Solving this, we have a
(y, t, c₁) = Cq.
(5)
(3 ) and (5) together constitute the general solution . At times it is desirable to replace c₁ in ( 5 ) by its value in terms of x and y. The solution is then
(x, y) = 1, \ 4 (y, t, $) = C₂.
Ex. 2.
dx xt
dy  dt yt
xy
813
dx dy = whence x y
Here we have
=
..x₁y ; and we have dt
dy =
yt
or c₁y dyt dt, cy
whence
4112  t2 = C29
or
xy  t2.= C2. .. The solution is
xây = 0 ,  t² = C2.
{ xy
3° It sometimes happens that we can find multipliers λ (x, y, t ) , u(x, y, t), v (x, y, t) such that, making use of the fact that dx = = dt = λαχ + μαν + vdt P e R XP + µQ + vR (a) this last member when combined with one of the others gives rise to an equation which can be solved ; or
DIFFERENTIAL EQUATIONS
156
(b) λP + µQ + vR may equal λ dx + µ dy + v dt = o
§ 65
zero, at the same time
satisfies the
condition
that
for integrability
(§ 35) ; or (c) by a choice of two sets of multipliers,
λ₁ dx + μ₁ dy + v₁ dt = λ2 dx + µ2 dy + v₂ dt λ₁P + μ₁Q + v₁R λqP + µyQ + v₂R
may be solvable . If we can find two independent relations by any of these methods, each involving an arbitrary constant, we have the general solution.
Ex. 3.
dt dx = dr_dy x y dx dy = we have x²  y² = C₁. x y
From
Letting λ = μ = 1 , v = 0 , we have
dx
dy
dt =
whence x + y = c₂t.
x + y dx =
Ex. 4.
cy  bt
dy at — CX
dt bx — ay
= v = c, we have by composition that the comLetting λ = a, μb, a dx + b dy + cdt mon ratio is equal to O .. a dx + b dy + c dt
o, whence ax + by + ct = C₁•
Similarly, letting λ = x, µ = y, v = t, we have x dx + y dy
Ex. 5 .
tdt = o, whence x² + y² + t² = €2.
dx dt = dy = 2 x² +y² 2 xy (x +y) i
SYSTEMS OF SIMULTANEOUS EQUATIONS
§ 66
I, V2 = 0, we have
Letting A = I , µ₁ = I , v₁ = o, and λ½ = 1 , μ₂ =
_dxdy , whence I dx + dy =  y)² x +y (x + y)² (x −
157
I x y
C1,
2y = c₁ (x² — y²).
or
dt
dx + dy =
Again
whence x + y = Cqt.
(x +y)²¯¯ (x + y) t '
Ex.
6.
_y dy  dt x dx_ y dy · yt xt y
Ex.
7.
dx  dy  dt x y I + 12
Ex.
8.
dx
= yt
Ex. 9.
Ex. 10.
dt • dy = xt x+y
dx x² — t² — y² 
dy 2 xy
dt 2 xt
dx  dy  dt t+ x x +y
y+ t
dx x² + y² + yt
dx
Ex. 12.
y³x − 2 x¹
=
dt
dy
=
Ex . 11.
=
x² + y² — xt dy 2 y¹ — x³y
(x + y)t
dt = 91(x³ — µ³)
66. Geometrical Interpretation.  P(x, y, z),* Q(x, y, z) , R(x, y, z) may be looked upon as determining the line X x Y  y_Z = s P R е through the point (x, y, z), which is any point in space. An integral dx  dy  dz curve ofthe system will be, then, a curve such that, at R P е * From now on we shall use the three letters x, y, z, instead of x, y, t.
158
DIFFERENTIAL EQUATIONS
§ 66
each point of it, it is tangent to the line at that point determined by P, Q, R.
The general solution, we have seen , consists of two rela=
tions, { (x, y = 4 } , involving two arbitrary constants . That is, y,, z) = represents a doubly infinite system of curves, solution general the
which are the intersections of two singly infinite systems of surfaces .* Thus in the case of Ex. 3 , § 65 , the integral curves are the intersections of the family of cylinders x²  y² = ₁ with the family of planes 40 we saw that a solution of the total In x + y ― C₂z = 0. differential equation Pdx + Qdy + Rdzo represents a surface such that, at each point (x, y, z) of it, it is tangent to the plane P(X − x) + Q( Y − y) + R(Z− 2) = o , the direction cosines of whose normal are proportional to P, Q, R. Hence we see that the intedy  dz dx = gral curves of cut orthogonally any integral surface of R P е PdxQdy Rdz = 0 . Since we can find a family of integral surfaces of Pdx + Q dy + R dz = 0 only when P, Q, R satisfy the condition for integrability [ $ 35 , ( 3 ) ] , we see that only in this case will there exist a family of surfaces of which the family of integral curves dx dy dz = = of are orthogonal trajectories. Thus, since yz dx + zx dy P е R
+ xy dz = o has xyz = c as its general solution , we see from Ex. 1 , § 65, that the curves of intersection of the cylinders x² — y² = c₁ and y² — z² = c₂ are cut orthogonally by the family of surfaces xyz = c. On the other hand, since xz dx + yzdy + xy dzo does not satisfy * Supposing P, Q, R single valued functions of x, y, z, there passes through any point (xo,yo, 30) the single curve { u (x, y, z) = u (xo, yo, 30) } , since the differential  v(x, y, z) = v (X0, Yo , %0) equations determine a single direction at each point in space. If P, Q, R are not all singlevalued functions, that is, if the differential equations are not both of the first degree, then more than one line (or direction) will correspond to a set of values of (x, y, z) , and there will be more than one integral curve passing through a point. In this case, u and v will not be singlevalued , that is, when the solutions are cleared of fractions and rationalized , the constants of integration do not enter to the first degree. This is analogous to what we found in the case of a single equation of the first order in two variables (§ 20).
$ 67
SYSTEMS OF SIMULTANEOUS EQUATIONS
159
the condition for integrability, there is no family of surfaces which is x c₁y = 0 cut orthogonally by the curves whose equations are xy32 = C2 (Ex. 2 , § 65) .
The converse problem of finding the orthogonal tra
jectories of a family of surfaces whose equation is f(x, y, z) = c is
21213
always possible, at least theoretically. For this necessitates solving the system dx dy dz = aƒ aƒ aƒ дх ду Əz Ex. Find the orthogonal trajectories of the family of surfaces xy = Cz. 67. Systems of Total Differential Equations. 
If we have two
total differential equations in three variables ,*
Pidx + Qidy + R₁dz = 0, (1) P₂ dx + Q₂ dy + R» dz = 0 , it can be proved (but the limits of this book will not permit our doing so here), that the general solution consists of two relations among the variables, involving two arbitrary constants. proceed as follows :
In actual practice we
If each of equations (1 ) separately satisfies the condition for integrability [ § 35 , ( 3 ) ] , we solve each one, and thus obtain the solution of our system . If only one of the equations satisfies the condition for integrability, we integrate that one, obtaining a relation
(x, y, z) = c .
Solving
this for one of the variables , we replace this and its derivative in the other equation by their values, thus giving rise to an equation in two variables only. Its solution, together with p(x, y, z) = c₁, already found, constitutes the solution of the system of equations. * The substance of this paragraph is at once applicable to the case of n equations in n + 1 variables.
DIFFERENTIAL EQUATIONS
160
§ 68
If neither of the equations is separately integrable, it is sometimes desirable to put ( 1 ) in the form
dx =dy = dz P R' е
(2)
2 R₂P₁, R = P₁Q2  P2Q1. P= Q1R₂ Q₂R₁ , Q = R₁P₂The methods of § 65 may now be tried.
If they do not work,
312 313
then taking one of the variables as the independent one, say z, the equations may be written. dx P = dz R' (3) dy = Q dz R
The general method of § 63 applies here. 68. Differential Equations of Higher Order than the First reducible to Systems of Equations of the First Order.  Given a single equation with one dependent variable. We may suppose it solved for the highest ordered derivative; thus, suppose we have the equation
(1)
If we put
d³y = dy ولاdx' dx3 /(x,,, 2, 23) 22 dy = ı dx1, and d2y_dy dx2 dx =12,
(1 ) may be replaced by the system of three equations of the first order 213
dy y1, dx = (2)
dy1 y dx = 2, dy2 =f(x, y, y₁, y2). dx
As an illustration,
d2y + y = 0 dx2
§ 69
SYSTEMS OF SIMULTANEOUS EQUATIONS
161
is equivalent to the system
dy dx =
y1
dy1 dx == y
, or dx_dy_dy1 . I 11 y
Using the last two terms, we have y² +y₁² = c12, or y₁ = √12  y².
dy = dx, or sin12 = x + €2,  2 C1 V₁₁21 whence
y = c1 sin (x + c2).
In an entirely analogous manner, a system of n equations of any order in n dependent variables may be replaced by a system of equations of the first order by letting each of the derivatives of the dependent variables up to the next to the highest ordered , in the case of each variable, be a new variable. Thus, by dx letting x1, the system of equations of § 64 may be written dt dx = x1, dt
dy = x1 + x  cost, dt dx1 — cost, = x1− 2 x + y + e²t dt or
dx x1
dy x1 + x COS t
dx1 x1−2x + y + e²¢
Cos t
dt I
69. Summary.  To solve a system of n ordinary differential equations involving n dependent variables we differentiate these equations a sufficient number of times to enable us to eliminate n — I of the dependent variables and all their derivatives, thus giving rise to a single equation involving only the remaining dependent variable . We integrate this, and substituting for this variable and its derivatives their values in terms of the independent variable in any n— 1 of the
162
DIFFERENTIAL EQUATIONS
§ 69
equations, we have a new system of n  I equations in n − 1 dependent variables. Repeating this, we find the value of a second variable and reduce the number of equations again, and so on. Or we can treat all of the dependent variables symmetrically by solving for each one separately, and then finding the relations among the constants of integration by substituting in some one of the original equations (§ 63). While this method is frequently not practicable , it can be carried out very readily in case the equations are linear with constant coefficients (§ 64). If the equations are of the first order, special methods can at times be resorted to (§ 65 ) . A system of n total differential equations in n + 1 variables can be written as a system of ordinary differential equations to which the methods of § 63 and § 65 apply ($ 67). A single differential equation in one dependent variable of higher order than the first, also a system of ŉ such in ŉ dependent variables may be replaced by a corresponding system of differential equations of the first order, to which at times the special methods of § 65 apply ($ 68). It is almost needless to add that if each of a system of equations
involves a single dependent variable, each is to be integrated independently of the others. Thus, see examples 1 , 2 , 3, 4 below. Ex. 1. Find the path traced out by a particle moving in a vacuum and acted upon by gravity only, if it is given an initial velocity v in a direction making an angle a with the horizontal plane . Ex. 2. If the particle in Ex. 1 moves in a medium which exerts a resisting force proportional to the velocity, find its path. Ex. 3. A particle moves about a center of attraction varying directly as the distance ; determine its motion, if it starts to move
$ 69
SYSTEMS OF SIMULTANEOUS EQUATIONS
163
from a point on the axis of x at a distance a from the center, and with an initial velocity 7% making an angle a with the axis of x. [ If the attracting force is P, and r is the distance of the particle from the the center of the force , the equations of motion are
d2x =
dt2 d'y = dt2
py. r
In this case P = k²r.] Ex. 4. If the force is a repulsive one, study the motion of the particle in Ex. 3. Ex. 5. A solid of revolution with one point of its axis of symmetry fixed, is acted upon by gravity only. Find its angular velocity and the position of the instantaneous axis of rotation in the body. [ If A, B, C are the moments of inertia of the body with respect to the principal axes of the momental ellipsoid about the fixed point, and p, q, r are the components of the angular velocity on those axes at any instant, Euler's equations are,
dp + ( C — A) qr = 0, Ah Α dt — A) rp = 0, − (C — A dq — dt
C
dr dt
= 0,
since B = A. ] Ex. 6. The component of the velocity of a particle parallel to each of the coördinate axes is proportional to the product of the other two coördinates .
Find its path, and the time of describing a
given portion in case the curve passes through the origin.
CHAPTER XI INTEGRATION IN SERIES 70. The Existence Theorem. 
The number of classes of differen
tial equations that can be integrated by quadratures or other purely elementary means is very small, compared with the number of possible classes of equations. In the General Theory of Differential Equations it is proved that every ordinary differential equation with one dependent variable (and every system of n equations with ŉ dependent variables) has a solution, in general, involving a definite number of arbitrary constants.
A proper understanding of the proof
of this theorem implies a knowledge of the Theory of Functions, which is not assumed here. A demonstration of the theorem will be found in almost any book dealing with the subject, presupposing a knowledge of at least the elements of the Theory of Functions.* dy 1° For an equation of the first order = F (x,y) ,† the theorem dx of existence of an integral is : * Cauchy ( 17891857) was the first to prove this theorem. In fact he gave two proofs of it, which have become classic. For a demonstration of this theorem a student familiar with the elements of the Theory of Functions may consult among other books, Murray, Differential Equations, p. 190 ; Schlesinger, Differentialgleichungen, Chapter I ; Picard , Traité d'Analyse, Vol. II , Chapter XI . More recently Picard (1856 ) gave another proof, which may be found in his Traité d'Analyse, Vol. II , p. 301, and Vol . III , p . 88, and also in the Bulletin of the New York Mathematical Society, Vol. I, pp. 12–16. † A differential equation of the first order√ f( x, y, y) =o may be supposed solved for (x,y,d (x, dy so that it takes the form dy = dx " dx F(x, y). 164
$ 70
INTEGRAT
ION
IN SERIES
165
* IfF(x,y) is finite, continuous, and singlevalued,
and has a finite
partial derivative with respect to y (see Picard , Vol . II , p . 292 ) , as long as x and y are restricted to certain regions, then if x and yo are a pair of values lying in these regions, we can find one integral y, and only one, which will take the value yo when x takes the value xv. In the proof of the theorem, y is found in the form of an infinite series ... ... Jo +1 (x − xo) + c₂ (x − xo)² + ··· + C₂ (x − xo)” + ·**, which series satisfies the equation when substituted in it for y, and besides is convergent for values of x sufficiently near to x .
By the
change of variable x = x  xo, the differential equation takes the
dy dx =F(x, y),
form
and the solution takes the form
y = Yo + c₂
+ Cq x² + ... ··· + C₁₂ " + ···
Since yo may be chosen arbitrarily (within certain limits, however) , we see that in the case of a differential equation of the first order, one arbitrary constant enters. Remark. The existence theorem gives a sufficient condition for an integral, and moreover, it gives a form in which the integral may be put. But this condition is not always necessary. Equations for which the conditions of the theorem are not fulfilled may have integrals. In general, but not necessarily always, such integrals will then not be developable by Taylor's theorem, or they will not be unique. A few simple examples will illustrate this :
dy dx
y x' where x becomes indeterminate for x = 0, y = 0, has the solution y = cx.
* Singlevalued is used in the broad sense here. Although F (x, y) may have several values for a single pair of values of x and y, it will be said to be singlevalued when x andy are restricted to certain regions if, having selected some one of its possible values for a pair of values of x and y in their respective regions, it will take a definite value for every pair of values of x and in their regions . Thus, while F= ± √x +y y has two values for every pair of values of x and y, if we select the value + √2 for x = 1 , y = 1 , F will have a definite value so long as x and y are restricted to regions where both are positive .
166
DIFFERENTIAL EQUATIONS
§ 70
Here y takes the value o for x = 0. It is expressed as a (finite) Taylor series, but it will be noted that is undetermined when we put y = 0 for x = 0 ; that is, unlike the cases coming under the existence theorem, there is an indefinite number of solutions satisfying the initial condition. Moreover, it should be noted that it is impossible to find a finite value for c that will enable us to assign a value to y other than o for x = 0. x +y dy_x Again, dx = + y where x x becomes indeterminate for x = 0, y = o, has the solution y = x log x + cx. Here y takes the value o for x = o. But it is not possible to express the integral in the form of a Taylor series in powers of x. In this case also we have an indefinite number of integrals for the one initial value o, and no integrals for any other initial value ofy. I + 2x = ∞, when x = 0, = + 2x 2 dy_1 has for solution y = Vc + x + x². dx y y y = o. In order that yo for x = 0, we must have c = 0. We have then the single solution y = √x + x². This is not developable by Taylor's theorem in powers of x however, although it may be developed in powers of √x. 2º If we have a system of n equations of the first order involving n dependent variables, we may suppose them solved for the derivatives of each of these variables :
dy ... dx = f1(x, y, z, •••, w),
l
dz ==ƒ2(x, y, z, •••, W), dx
dw aw = ƒn (x, y, z, ... , w). dx The general existence theorem says, in this case , that iffi, f2, ... ***fn are all regular, * as long as x, y, z, … , w remain in certain regions, then if xo, Yo, zo, ..., wo are in these regions, a single set offunctions y, z, ..., w can be found to satisfy the system of equations and to take the values yo, 20, •••, w。 respectively when x takes the value xo.
* For definition of regular see the second footnote, p. 203.
INTEGRATION IN SERIES
§ 70
167
In the proof of the theorem these are found in the form of the infinite series
... y = Yo + α₁(x − xo) + α2(x − x0)² + ··· + a„(x − xo)” + ··*, ... + b₁ (x z = zo + b₁(x − x0) + b₂(x − xo) ² + ··· − xo)" + •••, • w = wo + k₁(x − xo) + k₂(x − x0) ² + ··· + k„ (x − xo)” + ……, ··· , which series are convergent so long as x is sufficiently close to xo. As before, we can make the transformation = x  Xo, which will then give our series as power series in
.
be chosen arbitrarily (within certain limits) .
Here yo, 20, … , w。 may Hence we see that a
system of n equations of the first order with n dependent variables, has its general solution involving n arbitrary constants. We saw (§ 68) that an equation, with one dependent variable, of the nth order may be replaced by a system of n equations of the first order, hence it follows at once that the general solution of a differential In the way
equation ofthe nth order involves n arbitrary constants.
in which the equivalent system of equations of the first order is found, we see that we may choose for these arbitrary constants the values which the dependent variable and each of its derivatives up to the (n1 ) st take for a given value of the independent variable. Hence these values are, in general, at our disposal in any given problem. Geometrically, this means : Of the single infinity of integral curves of a differential equation of the first order and degree, a single curve passes through a given point. Of the double infinity of integral curves of a differential equation of the second order and first degree, a single curve passes through a given point, in a given direction. Of the triple infinity of integral curves of a differential equation of the third order and first degree, a single curve passes through a given point, in a given direction, and having a given curvature at that point.
DIFFERENTIAL EQUATIONS
168
71. Singular Solutions . 
871
In the existence theorem of the previous
section stress should be laid upon the fact that the existence of an dy integral of = F(x, y) is assured only as long as F (x, y) is finite, dx continuous, and singlevalued in the region of (xo, yo) . equation is given in the form dy f(x, y, y') = o, where y' = dx '
If, now, our
we know that, in general, y ' is expressible as a finite, continuous, and singlevalued function of x and y in the region of (xo, yo) , and takes a perfectly definite finite value y'o for x = xo, y = Yo• It can be shown * that this will be true as long as
af(x, y, y') 0. dy' af
But if
= 0, dy' then the expression for y' in terms of x and y ceases to be singlevalued in the region of (xo, yo) .
So that in the region of such values
for x and y the existence theorem does not assert the existence of a solution. As a matter of fact, a solution does not exist there in general .
For from
af
ƒ = 0,
=0 dy'
we can solve for y and y', thus
y = $(x), y' = 41(x) ; and only in exceptional cases will
$1 (x) ==
do(x) dx
* A proof of this theorem will be found in many works on Analysis ; for example see Liebmann, Lehrbuch der Differentialgleichungen, p . 8.
§ 72
INTEGRATIO
N
If it should happen that
IN SERIES
169
do(x) ₁(x) = dx 2 then y = p(x) is a solution
of the equation ; and since it is usually distinct from the general solution, it is a singular solution. Moreover, it is identical with the singular solution we encountered in Chapter V. Similarly in the case of differential equations of higher order than the first, singular solutions may occur. Thus if there is a solution of af also vanishes, this solution is, in f(x, y, y' , ..., y(x ) = o for which Jy(n) general, a singular one. * More generally, a system of equations of the first order dy Ji (x, y, 2, ... w, dx
dw = 0, dx
dy W, Jn Ja(x, y, 2, ... , dx
dw = 0, dx
(to which any system of m equations in m dependent variables is always reducible, § 68) may have singular solutions under certain. conditions . See Picard, Vol. III , p . 52 .
72. Integration, in Series, of an Equation of the First Order. – If the equation dy = F(x, y) (1) dx cannot be integrated by any of the known elementary methods, the existence theorem tells us that if F(x, y,) is finite, continuous, and singlevalued in the regions containing x = 0, y = co (there is no loss in assuming x = o ; since this amounts to presupposing the substitution = x  x。 to have been made, in case xo) , one and only one solution exists which takes the value of co for x = 0 . But this solu
* Liebmann, loc . cit. p. 113 ; Boole, Differential Equations, p. 229.
170
DIFFERENTIAL EQUATIONS
$ 72
tion is given by the existence theorem in the form of an infinite series. In actual practice we assume
(2)
y = n=0 Σ cnx" = co + C₁x + C₂x² + ... + Cnoch + ...
and substitute this in the differential equation ( 1 ) .
Then equating
coefficients, we may calculate as many of the 's in (2 ) as we please. The existence theorem vouches for the convergence of the series (2 ) . Three cases may arise : 1° No general law of the coefficients in ( 2) shows itself ; in this case we can only approximate the solution in actual practice. * O 2 A general law of the coefficients in (2 ) appears ; then we can write down the general term of the series, which is equivalent to say
ing that the whole series is known. 3° All the terms after a certain one may turn out to be zero ; in this case we have the solution in finite form.
Remark.  Cases 2° and 3° seldom occur except when the equation can be solved directly. So that this method of integrating equations of the first order is not of great practical importance for the mere purpose of integrating. But for theoretical purposes it is of the greatest importance. It may be noted that while every linear equation of the first order can be solved by quadratures, it is not always possible to perform these in terms of simple functions. In such cases this method, or that of § 74, will apply at times.
Ex. 1.
dy = x + y². dx
Here x + y² is finite, continuous, and singlevalued for all values of x and y. Put y = co + x + €²x² + ... * While the general existence theorem tells us that the series so obtained is convergent, as long as we restrict ourselves to proper values of the variables, the convergence may be slow for certain values so that the degree of approximation, even after having calculated a fairly large number of coefficients, may not be great. This will usually be true for values of the variables near any for which F (x, y) ceases to be finite, continuous, or singlevalued .
INTEGRATION IN SERIES
872
171
Substituting, we must have 21 G1 + 2 €qX + 3 Cgx² + ... ··· + n c₂x²  ¹ + ··· = x + (% + 4x + ···)³.
Equating coefficients, we have
91 = 602
..4 =
Co²,
2 C₂ = 2 Coc₁ +I 2
+ co³,
I =
3 C3 = 2 Coc₂ + G₁₂²
4 C4 = 2 CoC3 + 2 C162
co
+ co¹,
5 .. 4₁ = 12 √ co² + co³,
2 k C2x = 2 CoC2x1 + 2 C₁C2k  2 + ·
+ 2Ck1ck•
(2 k + 1 ) C2k+1 = 2 CoC2k + 2 C₁C2k 1 + ... + 2 Cx 1 Cx+1 + C .
Here each coefficient can be calculated in terms of the preceding ones, and consequently in terms of the single one ‹ . We can calculate as many as we please, but no general law shows itself. solution
y = co + co²x + ·( + co³) x² + (
Our
¢o + co *) x² + ...
extended as far as we care to calculate the coefficients, is only an approximation of the solution .
dy 2y = dx I x2°
Ex. 2.
Here I
2y x²2 is finite, continuous, and singlevalued in the region
of (o, co), where co is any value of y.
DIFFERENTIAL EQUATIONS
172
$ 72
Put yo + x + C»x² + ... + Cnoch + •••, and substitute in the equation, after having cleared of fractions. We must have (1 − x²) (c1 + 2 €qx C2X + 3 €3x² + ··· + nc₁, xn−¹ + ...) =
Equating coefficients, C1 = 2 Co. 2C,= 2=4 C 3 C3
C12 C₂ = 4 Co,
.. C₂ = 2 Co. .. C3 = 2 Co.
The general law seems to be c₂ = 2 Co.
We shall prove this to be
the case by showing that if it holds for ₂ it holds for n+1.
For,
(n + 1) C₂ +1 − (n − 1) cn1 = 2 Cn. Now
Cn1 = Cn = 2 Co. • . (n + 1 ) Cn+ 1 = 2 ( n − 1 ) co + 4 Co = 2 (n + 1 ) Co,
or
Cn+ 1 = 2 Co.
Hence the solution, in the form of an infinite series, is
y = co(1 + 2 x + 2 x² + 2 x² +
• + 2 x² + ···) .
[ This equation can be integrated by the methods of Chapter II . Let the student do this, and compare the result with the one here obtained.] dy When the equation dx F (x, y) is such that the various derivatives of F can be readily calculated numerically for special values of x and y, the following method is sometimes found practicable : We have seen that the solution given by the general existence theorem has the form
... y = cota + c₁ (x (x − −x xo) c₂ (x − xo)" + •••. y= ) ++ C₂(x − xo) ² + ··· + Cn
INTEGRATIO
N IN SERIES
$ 73
173
Using the general form of the Taylor development of a function, we see that when Co =yo, C1 =
dy dx
...
Cn = I (dry n ! don
we obtain that solution which takes the value y。 for x = ~0.
From the differential equation , we have dy C1 = d = F (xo, Yo) . xo
Differentiating the differential equation we have
day = F OF dy + dx2 dy dx' дх
whence
I d'y OF I OF I = C1 + C2 == 2 ! \ dx² 0 2 ! дх 0 2 ! ду 0
Differentiating again, we find c, in terms of 611 and C2. And so on. The student will find on applying this method to the examples above that it works very readily in the case of Ex. 1 .
73. Riccati's Equation. 
The equation studied by Count Riccati (16761754) , and to which his name has been given, is of the form
dy + by² = cxm, dx where b, c, m are constants.* type.
The equation in Ex. 1 , § 72 is of this
For certain special values of b, c, m this equation can be inte
dy * Frequently the equation x dx
ay+ By2:= yxn is taken as the type of a Riccati
equation. This is obviously reducible to the other by the transformation z = xa ,y = uz.
DIFFERENTIAL EQUATIONS
174
grated in finite terms.
$ 73
(See Ex. 4, § 18 ; also Forsyth, p . 170 ; Boole,
Chapter VI ; Johnson, Chapter IX . ) But in general, the only way to get the solution is to integrate in series. Riccati equations frequently arise, and it is often desirable to make use of the properties of their solutions without actually knowing the latter. On the other hand, it is sometimes possible to find the general solution by quadratures or by merely algebraic processes, when certain information is at hand. The following properties will at times prove of value : We shall consider a more general form , which is now usually considered as the type of a Riccati equation, dy  = X。 + X₁y + X₂y², dx where Xo, X1, X2 are functions of x or constants. 1° If a particular integral y, is known, the substitution y =
transforms the equation to
+11
dz + (X₁ + 2 y₁ X2)% = — X2, which is linear, dx
and can therefore be solved by two quadratures (§ 13) .
Hence
we have, if a particular integral y is known, the transformation I y == + y₁ gives rise to a linear equation in z which can be solved by 2
two quadratures . 2° Since the form of the solution of a linear equation of the first order is z = y(x) + Cd (x) , that is, the constant of integration enters linearly, we have that
I
y == +y₁ (x) ==
+11 (x) y( x ) + C8( x )
α(x) + CB(x) . y(x ) + C8( x )
$ 73
INTEGRATION IN SERIES
175
Hence, the constant of integration enters bilinearly in the general solution ofthe Riccati equation. α + BC 3° The equation y = may be looked upon as a bilinear 7 + 80
transformation of C into y, which latter is a particular solution as soon as a value of C is fixed . Corresponding to any four values of C, say C1, C2, C3, C4, we have y₁, y2, ys, 4. Since double ratios are left unaltered by bilinear transformations , we have { V1, V2, V39 Y4} = { C1, C2, C3, C4} = a constant. Hence, ify1, 12, Y'з, Y4 are any fourparticular integrals, the function (Y4 — Yз) ( Y½ — Y₁) is equal to a constantfor all values of x. (Y4 —Y1) (Y2 — Y3) 4° As a direct consequence of 3 ° it follows that if we know three particular integrals Y1, Y2, Ys, the general solution is given at once by
y— ys — = c Y2Ys ; y yi Y2 — Y3)  Y1) — CY₁ (Y₂ y³) , i.e. y is given by purely algewhence y = Y3 (Y2 — Ya Y11213) braic means. 5° If y, and y½ are two known particular integrals, put z = and take the derivative of the logarithm of both sides. I dz z dx Since
dy I y y₁ dx
dy dx
dy 2 = X + X₁y + X₂y², dx
dy dx = Xo + X19₁ + X2 Y₁², dy2 = X + X₁Y2 + X2y²², dx
I
y  y2 This gives
(dy  dy2 dx y y2 dx
DIFFERENTIAL EQUATIONS
176
$ 73
I dz we have z dx
z = CesX2(V₁−V₂)dx,
whence
i.e.
= X2 (J1  J₂) ;
— CeSX2(V1—Y2)dx, Y_— Y1 = y y2
from which y can be gotten at once.
Hence, if two particular inte
grals y, and y₂ are known, the transformation z = YY₁ leads to an y y2 equation in which the variables are separated, so that it can be solved by a single quadrature. Properties 1 °, 4º, 5º call attention to the close analogy of the Riccati equation to linear differential equations ; for the knowledge of each additional particular integral brings us nearer to the general solution .
Thus, the knowledge of a single particular integral enables
us to reduce the problem of solving a Riccati equation to one of solving a linear equation of the first order, that is to one involving two quadratures; the knowledge of two particular integrals enables us to find the general solution by performing a single quadrature ; while a knowledge ofthree particular integrals gives us the general solution by a very simple algebraic process. 6° As a matter of fact, the substitution
I
dz
y= X₂z dx transforms the Riccati equation into the homogeneous linear equation of the second order
dz d2% X2 − ( X₁X2 + X'2)·dx + XoX²2 % = 0, dx²
where X ', = dX2 . dx
INTEGRATION IN SERIES
$ 74
177
[ Let the student show that, conversely, the substitution y = eſedz transforms a homogeneous linear equation of the second order into a Riccati equation. ]
74. Integration, in Series, of Equations of Higher Order than the First. tive *
If the equation when solved for the highest ordered deriva
day dy day = F ( x, y, dx dx3 ' dx2
is such that the successive derivatives of F can be readily calculated numerically for special values of
x, y,
dy d'y dx' dx²'
a method analogous to that given at the end of § 72 may be employed. The solution given by the general existence theorem being in the form y = co + c₁(x − xo) + €2(x − xo)² + ... ··· + Cn (x − xo) " + ···, we know from the general form of the Taylor development of a function that I ' d'y dy I (day) ... C2 = 2 ! dx² Co = yo, C1 = dx n ! dan
Now the general solution involves three arbitrary constants, and we saw (§ 70, at the end) that a particular solution will be determined as soon as we fix the values of y, dy d'y for xx ; let us call them ' dx2 dx Yo, Yo', Yo". * To fix the ideas we shall illustrate with an equation of the third order, although this method, when practicable, applies to equations of any order.
178
DIFFERENTIAL EQUATIONS
$ 74
From the differential equation we have at once
=
I (day C3 == 3 3! dxo
I ¡F (xo, Yo, Yo' , Yo" ) . 3!
Differentiating the equation we can calculate
xo, yo,
dy 'd²y dx, o dx
'd'y dx
in terms of
(d³y\ ; dr
so that we can find c, in terms of co, C1, C2, C3.
And so on.
When this method does not work readily, we may employ the first method of § 72 as there given .
But again, only in comparatively few
cases will this turn out to be practicable .
The following modification
of this method has been found of special service in the case of linear differential equations in which the coefficients are polynomials in x, and such that when we substitute x" for y in the lefthand member, there results only a small number of distinct powers of x (preferably, not more than two).
In the case of a Cauchy equation ( § 51 ) there
results a single power of x, and the equation can be solved by purely algebraic means. *
If we take the equation day — xy = 0, dx2 the result of putting y = x" in the left hand member is
m (m  1) xm2  x +1. * Thus, putting y = xm in the left hand member of ( 1) , § 51 , we get [k¸ m (m − 1) ….. (m − n + 1) + k¸ m (m − 1) ….. (m·一九 +2) + ... + kn 1 m + kn] xm. Equating the coefficient of xm to zero and solving for m, we get in general n distinct particular solutions and therefore the complementary function. It is readily seen to be the same as that found in § 51. The cases of equal and complex values of m can be treated entirely analogously to those for linear equations with constant coefficients. In some respects, the Cauchy equation is simpler than the corresponding linear equation with constant coefficients. But for actually obtaining its solution, especially when there is a righthand member, it is usually simpler to transform it to a linear equation with constant coefficients, as was done in § 51 .
$ 74
INTEGRATION IN SERIES
179
Here there are two distinct powers of x which differ by 3, and m  2 is the smaller exponent.
Hence, if we let
y = cox + c₁xm+3 + C₂xm+6 + ... + Gam+ar + ...
we shall have a solution if 1° m is so chosen that m (m − 1 ) = 0 , i.e. m = 0 or 1 , 2° the c's are so chosen that the rest of the terms cancel each other in pairs, i.e. we must have
(m + 3) (m + 2) 1 (m + 6) (m +5)
Co
= 0,
1
= 0,
(m + 3 r) (m + 3 r − 1 ) c, — Cr −1 = 0,
I ·Cr1• (m + 3 r) ( m + 3 r − 1) I For mo, we have c₁ == Co 3.2
I C1 C2 =6.5
=
= I.4 Co 6!
I C2
=
I 12. II
=
Cg = 9.8
C =
Co
3!
I • 4.7 Co 9! I. 4.7 · IO Co
12 !
... = 1 · 4 · 7 ··· [ 1 + 3 (r − 1) ] con (3 r) !
DIFFERENTIAL EQUATIONS
180 I
+ 3!
$ 74
+ 1 · 4x + 1 · 4 · 7x + ... 6! 9! +
is an integral . constant.
I. 4 · ... 7 ··· [ 1 + 3 (r− 1 )] x³r +· +...) (3r) !
Let us call it Ay , where A, like co, is an arbitrary
2
I ==
For m = 1 we have c₁
Co =
4.3
C2 =
Co, 4!
I 41 = 2.5 Co 7!
7.6
I 2.5.8 Co C2= 10.9 IO !
C₁ =
+· •. Cox ( I1+ +23234x³+
is also an integral. stant.
y and y
2.5 7!
... 2 · 5 · 8 ··· [ 2 + 3 ( r− 1) ] co. (1 + 3r) !
... + 2 · 5 · 8 ··· [ 2 + 3 ( r− 1 )] An³ , r+ ...) (1 +37) !
Let us call it By2, where B is an arbitrary conare obviously linearly independent. Moreover,
they are convergent by the general existence theorem.
Hence,
y= Ay₁ + By, is the general solution, since it contains two arbitrary constants. If the righthand member of the equation had not been zero, we would have proceeded to find the particular integral by a similar method.
Thus, suppose the equation had been
day ― ·xy = 2 x³. dx2
181
INTEGRATION IN SERIES
$ 74
Since the result of putting y = com in the lefthand member is
— Coxm+1, c。m (m − 1)xm –2 Conti + G₁₂ +3 + ... + cxcm +3 + ... will be a particular integral, prom 2 vided cm (m— (m − 1 ) xm– = 2 x³, and the other terms that arise destroy each other in pairs. The first of these will be true if
and
m2 =
3,
... m = — I,
com (m − 1 ) ==
2,
•• Co =
I.
The other terms will destroy each other in pairs if
€1 =
= I 2!
I Co 2. I I
C2 =
C1
.. C2 = 3 5!
C2
.. C3 
I =
8.7
3.6 8!
Cr1´´` · * . c, = 3 •· 6 · 9 ... ··· 3 (r − 1) ¸ (31 ) ! (3 r  1) (32) I
Hence a particular integral is I 3 x² + 25+ ... + 3.6.93 (  1 ) 231+ .. 2! 5! (3r 1)! The above example suggests a general method,* in case the result of substituting y = x" in the left hand member of the equation gives * As mentioned in the Remark, § 72, this method applies to linear equations of the first order as well as to those of higher orders.
DIFFERENTIAL EQUATIONS
182
$ 74
rise to only two distinct powers of x, say f(m) x² + $ ( m) xht!, where is a positive integer. By the manner in which ƒ(m) and (m) arise , it is obvious that one of them at least must be of degree n in m. I. To find the complementary function we proceed as follows : (a) Suppose f(m ) is of degree n in m , so that ƒ(m) = o has n roots, M1, M2, M3, ... ••• M,, all distinct, or some repeated . time + Gam+1 + ... + C₁xm+rl + ..., we have, on substiLetting y tuting in the lefthand member of the differential equation,
cof(m) x + cop (m) x²+1 + c₁f(m + 1) xch+1 + c₁$ (m + 1 ) xch+2 + c₂f(m + 2 l) xh + 21 + c₂Þ (m + 2 7) xh+31 + + cr_1f(m + [ r − 1 ] 7)xch+(r− 1)² + Cr_1 $ (m + [ r− 1 ] )
+r
+ cf(m + rl) x²+r + c,$ (m + rl) xch+(x+ 1)? +
This will be zero provided 1° ƒ(m) = 0 , i.e. m = m1, M2, *** , Mn ; r = 1 , 2, 3, • * •, ∞∞ 2° c‚f(m + rl) + c, 18 ( m + [ r − 1 ] ) = 0, for
m = m1, M2, *** , Mn
or
== $(m + [ r− 1 ] ¹) cr1 = f(m + rl) ... = ( − 1) , $(m + [r− 1 ] Z) $ (m + [ r − 2] /) ··· þ(m + Z) $(m) Co. f(m + rl)f (m + [ r − 1 ] l) ... ···ƒ (m + 2 l )ƒ ( m + 2)
* An analogous method can be deduced in the case of a larger number, but this will not be done here.
INTEGRATIO
$ 74
N
IN SERIES
183
To each value of m corresponds, then, in general, a particular integral. * If any c vanishes, all that follow do so, and that integral appears in finite form. If two values of m are equal, of course, the same particular integral will correspond to these.
Moreover, if two of the m's differ by an
integral multiple of 1, say m = m₁ +gl, then corresponding to the , will be infinite [ since f(m²) =ƒ(m₁ +gl) = o ] , unless the numerator is also zero . Hence the method here given gives us only as many particular solutions in general, as there are distinct roots of ƒ ( m ) = o, whose differences
smaller value m₁, the coefficient
are not multiples of 7. The remaining solutions must then be sought by a modification of our process.† If f (m) is of lower degree than n, while the above method may lead to infinite series which will satisfy the equation, the general theory gives us no assurance that they are convergent. So that, unless the solutions come out in finite form, it is best to make use of the fact that if f(m) is not of the degree n, (b) If (m) is of degree n in m , values of m, say m ' , m'2, ... m'n'
(m) is.
( m) = o will be satisfied by n
Letting y = com + .c_.1xm 1 + c_2xm21 + ...·+ c_rxm−rl + •••, we have on substituting in the lefthand member of the equation,
Co (m)x +ε tap(m − n
+ cof(m)x + c_1f(m  1)xch1
+c_2 $(m −27)x²¹ + c_₂f(m −21)x  2
+ * It should be noted that corresponding to any m which is not a positive integer, we have a solution which is not a power series, but xm multiplied by a power series. Although the general existence theorem no longer applies here, because the coefficients in the equation, when the leading coefficient is made unity, cease to be finite for x = 0, the general theory of linear equations assures us of the convergence of the series for certain values of x. (Schlesinger, Differentialgleichungen, § 24.) In this case the particular solutions, which the general method fails to give, usually involve logarithmic terms. Thus see Ex. 2.
DIFFERENTIAL EQUATIONS
184
$ 74
 [r− 1 ] 7) xh−(r−2)² + C_r+1ƒ(m − [ r − 1 ] /) xh−(r−1)l + C_r+1 $ (m − +c_, p(m  rl)x−(r−1)² + c_ ,ƒ (m — rl)xch—n +
This will be zero if 1°
(m) = 0, i.e. m = m'1, m'2, •• *, m'n,
2° Cr
f( m [r 1 ] ) cr+1 for (m  rl)
r = 1 , 2, 3, *** , ∞ m = m'₁, m'2, ..., m'n }
ƒ(m − [r − 1 ] l) ƒ ( m − [ r − 2 ] !) ··· ƒ(m − 1) ƒ(m) Co• == (( − 1)" 1)″ . $(m − rl)p(m — [ r − 1 ] 7) ….. $(m − 2 1) p (m − 1) To each value of m corresponds, in general, a particular integral rl 21 (co + c _1 x² + C_2 x¯² + ·... + C_, x¯¹² + •••). of the form Here is multiplied by a power series going according to negaThe latter reduces at once to an tive ascending powers of x. I ordinary power series in t if we put t = Of course, if any of the x roots of $( m) = o are repeated , or differ by a multiple of 7, the number of integrals obtained by this method will be less than n. It should be noted that the integrals found by methods (a) and (b) in case both f (m) and (m) are of the nth degree , are not distinct. In general (but not necessarily) , they are different in form, but only n of the functions defined by them can be linearly independent. * * An infinite series defines a function for those values of the variable only for which the series is convergent. The series found by methods (a) and (b) , being developments in the region of the origin and of ∞ respectively, may not, and frequently do not, converge for the same values of x. Hence if the series are infinite, it is usually impossible to compare them . But if the functions represented by each set can be " continued " into the region of the other, then a linear relation will be found to exist among any n + 1 of them.
INTEGRATIO
§ 74
N
IN SERIES
185
II. To find the particular integral in case the right hand member is a power of x, say Axs, we proceed as follows : If f(m) is of degree n, we must have, using the results found in
connection with method (a) above : 1° hs. Since h is a linear function of m, this will determine a single value of m, say m¸.
2° cof(m ) = A. 3° The remaining coefficients are determined as in (a) except that now m , is used for m. This method will be in default when ƒ(m¸) = o . Or using the results found in connection with method (b) *, we get the solution by putting : 1° h + l = s.
This will determine a value m ' of m.
2° cop(m¸') = A. 3° The remaining coefficients are determined as in (b) except that now m,' is used for m. If (m .') o at the same time that ƒ(m, ) = o , the particular integral will not be of the form here sought. Other means will have to be used to find it. For a perfectly general method for finding the particular integral see Schlesinger, § 54 .
Ex. 1.
d'y  dy 3 + 2y = x + 3xª. (x − x²) dx2 dx
Substituting a for y in the lefthand member, we have 1 m(m − 4)xm−¹ — [ m(m − 1 ) — 2 ]xm.
*Whether (m) is of degree n or not. The objection to using method. (a) or (b) in case f(m) or (m) is not of degree n is, that if the solution comes out in the form of an infinite series, the latter need not be convergent. If the solution appears in finite form , the method applies perfectly well ; if not, the convergence of the series must be looked into. In the case of the example worked out above, if the righthand member had been x, method (a) would have given the particular integral in the form of an infinite series, which turns out to be y₁1. On the other hand, method (b) gives I for the particular integral at once.
DIFFERENTIAL EQUATIONS
186
$ 74
../= 1, f (m) = m(m − 4) , (m) = − (m + 1 ) (m − 2) . Since ƒ (m) is of degree 2 , we shall use method (a) . Then mo or 4,
and
 p(m + [r− 1 ]) cr 1 = m + r  3c,1° −4 + m r f(m + r)
For mo we have
2 G₂ = 13co = Co, C1 I 4 3
C₂ = 23 c₁ == Cor 2 4 3 c3 = 3—3 c₂ = 0 , 34
= 0,
C4=
..
+ ·x + 3 3
or A(3 + 2x + x2) is an integral.
it Ayr For m
C₁ =
I+ I Co = 2 Co, I
C2:
2+ I C1 = 3 Co 2
4 we have
= C3 =
3+ 1 C₂ = 4 Co, 3
6₁ = 4 + 1 C3 = 5 Co,
Let us call
$ 74
INTEGRATION IN SERIES
.. Co(x² + 2x + 3 x² + ... ··· + nx² ++ ... ) is an integral. call it By2.
187 Let us
Therefore Ay₁ + By, is the complementary function . Since ( m ) is of the second degree, we can also use method (6) . Then m 2 or I,
and
For m
m  r  3c_r+1. ƒ (m — −1 − [r — c_, — — 1]) 6 r+ 1 = m r  2 (m  r) 2 we have
C1
II Co = 2 Co,  I
C_2 = C2:
I 2c_1 = 3 Co, ― 2
C3:
3 I C_2 = 4 Co, 3 ·
+n n x³n + ...) is an integral. ··· + .. Co(x² + 2x + 3 + 4x¹ + 5x² + ...
Let us call it A'y'. I we have For m
C1 =
500 I 4 9 Co = I 3 4
C_2 =
24,C 1 = 6 Co, 2 4 3
C_3 =
—3—4 C 33
= 7 Cos 4
n (½ +3 ) x¯” + ...) is an integral. · ·21 (4 x² + 5 x² + 6 x³ + ... + (n 4 Let us call it B'y₂'. Hence A'y ' + B'y ' is the complementary function. Here it turns out that y₁' — y₂ = Yr.
DIFFERENTIAL EQUATIONS
188
$.74
To find a particular integral, we consider each term of the righthand member separately.
For the term x we have, putting c。m (m − 4) xm −1 = x, m2 and co:= 
4
For m = 2
r I Cr1. r 2
Cr
..₁ = o , and all that follow are zero. Hence 
I is a particular integral corresponding to x.
4 Corresponding to 3 x¹, we have, putting
c。m (m − 4 ) xm − 1 = 3 x4, m = 5 and co
3 5
r+2
For m = 5
Cr
Cr1. r+ I
 Co • =3 29 = 26 IO · 3, 3 • a= 1q = ΙΟ 4, 3 = 6243.5 ΙΟ 5,
6 Cg =
Hence
3 · 6, ΙΟ
3 ... 4x² x² + ··· + (n + 1 ) x* +* + ···] ΙΟ [2x + 3x + 4
INTEGRATION IN SERIES
$ 74
189
is a particular integral corresponding to 3 x . Comparing this with 3  x¹) . y2 above, we see that it is equal to ΙΟ·(1½ — 3 Hence a particular integral is ΙΟ The complete solution is then I 2 y = Ay₁ + By½ — — x² — ΙΟ 3x4 . 4
Ex. 2. x
d'y dy + dx2 dx + y = 0.
The result of putting y = x" in the lefthand member is
m²xm1 + xm
../= 1, (m) = 1 .
f(m) = m²,
Hence method (a) applies .
m = 0, 0, I
I
Cr.
Cr =
=―
f(m + r) since mo is the only choice for m. I
Co = 
..1 =
I I 2 C1 =220
C2:
I
I
C3

x
x2 +
+ (21)2
is an integral.
(3 !) 2
In + + (− 1)"; (n !)2
Let us call it cy₁.
* This form could have been obtained at once by equating cop (m) xm to 3x4,
DIFFERENTIAL EQUATIONS
190
$ 74
To find a second integral, let y = y₁v + w , where v and w are two functions still to be determined . Since the single requirement of having this value of y satisfy the equation is imposed upon these two functions, a second relation between them may be assumed.
We
shall do this so as to simplify our work as much as possible. Substituting this value of y in the equation, and remembering that y₁ is an integral, we get
x
d2v d2w dw dv dy, dv = 0. +2 x + w +Y₁ x² + + dx dx dx dx2 dx dx2
Assume now that
x
d2v dv + dx dx2
(this is the second relation at our disposal).
Then
v = A +Blog x.
And our equation to determine w is
x
dew , dw + + w = 2B dy dx2 dx dx
= 2 B
Letting w
2 Bx 2 Bx2 + 2! 213 !
2 Bxs
2 Bxn + ... + (− 1)"; + 314 ! n ! (n + 1) !
co + G₁x + C₂x² + ... +
спость + oog
dw = C₁ + 2 C₂x + 3 Cgxx² + ... + (n + 1 ) Cn+¹xn + dx
whence
x
d2w dx2
2 Cgx + 6C3x² + ... + n (n + 1) cn+¹² + ···
INTEGRATIO
§ 74
N
IN SERIES
191
we have on equating coefficients Co +
1 = 2B,
G₂ + 46₂ = _2B 2! 2B
C₂ + 9C3 =
2 ! 3 !' 2B Cn + (n + 1 )²Cn + 1 = ( − 1) ” n ! (n + 1 ) ! '
Since we are looking for as simple a particular integral as possible, = ; then there will be no loss in assuming coo
G₁ = 2B, I (2 ¹² !)2 ( 1 + 1) 2 B ,
C2 =
I C3 = 1 ) 2 B. = (371) ² ( 1 + 22. + 3/
I En = ( − 1 )n +1¸ (n !)
I+ 2
I +=+ 3
+
2 B.
Hence the general solution is
x
xn ... + ( − 1)" ; + ... + ··· (3 !)² (n !)²
x2
X3
y = (A + Blog x) I
(11)² + (2 !)² x
x2
+2 B
+ !)²
(2 !)
+
13 I += + !)2
xn + ( 1) +1 ' (~ ) ( (n 1 !)2
+ ...
I I ... + 2 + 3 + ++1) +
]
DIFFERENTIAL EQUATIONS
192
Ex. 3. (x  x²)
$ 75
dy d'y +4 + 2y = 0. dx dx²
Ex . 4. 2 x2 d²y  xdy + (1 − x²)y = x². dx dx2
dy Ex. 5. (x − x²) dy +3 + 2y = 0. dx dx2
75. Gauss's Equation . of the equation
Hypergeometric Series.
The integration
d2y dy (Az² + Bz + C) dz2 1²x² + (Dz + E) dz 2 + Fy = 0, where A, B, C, D, E, F are constants and B²  4 AC0, leads to
a remarkable series exhaustively studied by Karl Friedrich Gauss This series and its differential equation were dis( 17771855 ) . covered by Euler ( 17071783 ) .
Putting z = ax + b, we have
[ Aa²x² + (2 Aab + Ba)x + Ab² + Bb + C] d² + (D'x + E') dy dx2 dx
+ Fly = 0, where D' , E', F' are constants. Choosing a and b so that Ab² + Bb + C = 0
and
2 Ab + B Aao,
and dividing by the coefficient of (x²  x)
d'y we have dx2'
dy + Ry = 0, (x² −  x) d²² + (Px + Q) dx2 dx where P, Q, R are constants.
INTEGRATIO
N IN SERIES
§ 75
193
Substituting " for y in the lefthand member, we have
1 — m(m − 1 − Q)xm−¹ + [ m² − ( 1 − P)m + R] x™. Putting Qy, 1 − P = − (α + B) , R = αß,
our equation takes the form (x²  x) d²y + [ (α + B + 1)x  • y] 7] d² + aẞy = 0, * dx2 dx
and the result of substituting y = xm in the lefthand member is ƒ(m)xm−¹ + $(m)x™,
where ƒ(m) =
m (m − 1 + y) , $ (m) = ( m + x) ( m + B) .
Using the method (a) of § 74, we have mo or I
Cr=
c p(m + r − 1) Cr1 = ( m + r + a − 1) ( m + r + B − 1 ) cr 1 • (m + r)(m + r + y− 1) f(m + r)
For mo, we have (1 =
C2 =
cn =
=α ·β I.·Y (α + 1 ) (B + 1) C1 = α( α + 1 ) ß(ẞ + 1 ) Co I 1.2.y (y + 1) 2 (x + 1)
3
¸α(α + 1 ) ... ··· ( α + n − 1 )ß ( B + 1 ) … .. (B + n − 1) Cos 1.2.3 ... n ・ y (y + 1 ) ·... ·· (y + n — 1 )
* This is usually referred to as Gauss's equation.
DIFFERENTIAL EQUATIONS
194 Putting
J₁ = I +
§ 75
= 1, we have the particular integral
œ(œ + 1 ) ẞ(B + 1 ) α.B ·x²+•... ·x + I. γ I · 2 ·. Y (y + 1 )
... ) ¸2n + + a(a + 1) ··· ( a + n − 1 ) B ( B + 1 ) ... (B + n − 1I) 1.2.3 . n . y (y + 1 ) ··· (y + n − 1)
This is the hypergeometric series, and is usually represented by
F(α, B, y, x). For m = 1  y, the student should show that the integral is co Y F(α  y + 1, B  y + 1 , 2  y, x). where y = xy If a or ẞ is a negative integer, while y is not, y₁ reduces to a polynomial. If y is a negative integer (including zero) , say y = go, while neither a nor ẞ is one of the integers from g to o, the coefficients in y₁ beginning with the gth become infinite. So that this form for y cannot be used.
(Another form involving log x can be found,
which together with y will give the general solution . ) If y= = go , where g is an integer , and a or ẞ is one of the in the (except intege from g to o, y, reduce to a polyno rs ing mial s and case when a or ß = y ; in this case a factor in the numera tor t with he gth b of every coeffic one in the denom inato ient eginning r o e n f z t h ; v term anish ach ther , and the ere hese ero actors eutrali ze out these zero factors gives us an availab b st result obtain le ed y riking roots of ƒ (m ) = 0 t t form for y₁ . *) In this case , althou gh he wo genera i t e t differ by an intege , no logari l r thmic erm nters n he soluti . on If γ is a positive number, say y = g > o, and neither a nor ẞ is one of the integers from 1 to g, the coefficients in y, become infinite. In this case a new form for y, involving log x can be found .
* See Schlesinger, Differentialgleichungen, § 34.
$ 75
INTEGRATION IN SERIES
195
If, however, y = g > o where g is an integer, and a or ẞ is one of the integers from 1 to g, y½ reduces to a polynomial (with the exceptional case of a or ß = y, which is handled in the same manner as the exceptional case for y a negative integer or zero) .
So that in this case
also no logarithmic term enters in the general solution . Using the method (b) of § 74, let the student show that
I y' = x—a F ( a, 1 + a −y, 1 + a − ß, 1),
I Y2' = XB‚F( ß, 1 + ß −y, 1 + ß − α,
)
are a pair of linearly independent integrals. If y₁ and y, are infinite series with finite coefficients, they will be convergent for all values of x less than 1 in absolute value, and divergent for all values of x greater than 1 in absolute value ; while y₁' and y ' are convergent as long as x is greater than I in absolute value, and divergent as long as x is less than 1 in absolute value. The hypergeometric series may at times represent wellknown functions. Thus let the student show that Ex. 1. F ( n, ß, ß, — − x) = ( 1 + x)" for ẞ any constant. Ex. 2. xF( 1 , 1 , 2 , −x) = log ( 1 + x).
Ex. 3. LimitB=00 F F ( 11,, B, 1,14 ) =
.
x2 = sin x. Ex. 4. Limit a, B=x xF( α, B, 3 , 2  4αB Ex. 5. Express as hypergeometric series the following functions :
I —— ₂ ( 1 + x) " + (1 − x) ”, ( 1 + x) " − ( 1 − x) ", cos x, e* + e˜³ . IX For further examples see Gauss, Collected Works, Vol. III , p. 127
CHAPTER XII PARTIAL DIFFERENTIAL EQUATIONS 76. Primitives involving Arbitrary Constants. ― Partial differential equations may be obtained from primitives involving either arbitrary constants or arbitrary functions.
Thus, consider the family of
spheres of fixed radius R, with their centers lying in the plane of 2 = 0.
(1)
The equation of this family is obviously
 a)² + (y — b) ² + z² = R², (x −
where a and b are arbitrary constants.
We shall consider x and y as independent variables, and shall put
22% 22% Əz 22% az = = S, = t. = 9, = r, == dxdy dx² Әх მ y2
Differentiating ( 1) with respect to x and y respectively, we get
(2)
x a + zp = 0,
(3)
y  b + zq = 0.
Eliminating a and b from ( 1 ) , ( 2) , (3) we get (4)
z²(p² + q² + 1 ) = R²,
which is known as the differential equation corresponding to the primi tive (1 ) ; on the other hand ( 1 ) is said to be a solution of (4). 196
$ 76
PARTIAL DIFFERENTIAL EQUATIONS
197
Perfectly generally, if we start with any relation involving two arbitrary constants and three variables, of which two are independent, (1)
(x, y, z, a, b) = 0,
where z is taken as the dependent variable, we get on differentiating with respect to x, and then to y
(2)
аф փ = 24+p +1 as дх аф
(3) ду
+9
= 0
Əz
We now have three equations from which to eliminate a and b. Doing this, there results
(4)
f(x, y, z, p, q) = 0,
a differential equation of the first order which has ( 1 ) for its primitive . If the primitive involves more than two arbitrary constants, more than three equations will be necessary to eliminate these, so that if only two independent variables are involved , the resulting differential equation will be of higher order than the first. Thus consider (5)
ax² + by² + cz² = 1 .
Differentiating with respect to x and y respectively, we have
(6)
ax + czp = 0,
(7)
by + czq = 0.
We have now only three equations from which to eliminate a, b, c. Hence we must differentiate again . to x we have
(8)
Differentiating (6) with respect
a + c(zr + p²) = 0.
198
DIFFERENTIAL EQUATIONS
$ 76
Eliminating a, b, c from (5) , (6) , ( 7) , (8) , we have
I x
(9)
O
zp
y O
zq
= 0, or
zr +p²
(xzr + xp²  zp) = 0,
which is a differential equation having ( 5 ) for its primitive .
Had we
differentiated (6) with respect to y, we would have gotten
(8')
czs + cpq = 0 ;
whence, on eliminating a, b, c from (5 ) , ( 6) , ( 7) , (8 ') ,
(9')
zs + pq = 0,
which is also a differential equation having (5) for its primitive.
Again,
differentiating (7) with respect to y, we get
(8")
b + c(zt + q ) = 0.
Eliminating a, b, c from (5 ) , (6) , ( 7) , (8 ") , we have (9")
yzt +yq² — zq = 0,
which is also a differential equation having (5 ) for its primitive. Since (9) , (9 ') , (9 ″ ) are all of the second order, there is no choice among them, and all may be said to be differential equations belonging to (5)* .
* Attention should be called to the fact that there is no such ambiguity in the case of the equation of the first order belonging to a primitive involving two constants. Since there are k + 1 kth derivatives of z depending on the number of times we differentiate with respect to x and to y, it is clear that, in order that a primitive should give rise to a unique partial differential equation of the nth order, the number of arbitrary constants it contains must be equal to the sum of all the integers from 2 to n + 1, i.e. n(n + 3) . 2
§ 77
PARTIAL DIFFERENTIAL EQUATIONS
199
Form the differential equations having the following equations for primitives, a, b, c being the arbitrary constants to be eliminated : Ex. 1. (x − a)² + (y − a)² + z² = b². Ex. 2. a(x² + 1²) + b²² = 1 . Ex. 3. z = ax + by + √a² + b².
Ex. 4. z = (x + a)(y + b). Ex. 5. (x− a)² + (y − b) ² + (≈ − c )² = 1. Ex. 6. z = ax + by + cxy. Ex. 7. ax + by + cz = 1.
77. Primitives involving Arbitrary Functions.
Let u and v be
two known functions of x, y, z, and suppose we have the arbitrary relation
(1)
(u, v) = 0.
Differentiating with respect to x and y respectively, we have
(2)
du მიმა дф/ди მა + Oz дидх + dz1)+ dv dx
(3)
dv du дф/ди მიმა 9+ 9 = 0. + диду + Əz 9 ) до ду
= 0,
In order that (2) and ( 3) may hold simultaneously, we must have
du  du ·p Эх + Əz
dv + dx
(4)
dv ƏzP
du 9 dz dy Ov = 0, or dv + 'q მე ду du
Pp + Qq = R,
+
DIFFERENTIAL EQUATIONS
200
§ 77
where
du dz P=
 მა dz
ότι ду dv " ду
du Эх
du dz
l:= მა дх
dv
Ju R = ду dv
მz 
 ду
Ju ax av
ax
This is a linear partial differential equation of the first order.
By
a linear partial differential equation of the first order we mean one that is linear in the derivatives of the dependent variable (the way in which the dependent variable itself enters playing no rôle). * If u and v are two known functions of x, y, z, and we have (1)
ƒ[x, y, z, $(u), 4 (v) ] = 0,
where ƒ is some known function, but & and ↓ are arbitrary functions, then, on differentiating, we have
(2)
(3)
du du aƒ + afp + af af аф '(u) дх + Iz дх Əz
= 0,
+
)
+
)
dv du Ju af af af+ of Չ = 0. + 9+ '(u) 9+ )( + 2 4( dy Əz dy dz Эф dy öz 24
Since (2 ) and (3 ) introduce two new functions p' (u), ¥ ' (v) , five equations are necessary to eliminate the four functions (u) , 4(v), p'(u), 4' (v). We must differentiate ( 2 ) and (3 ) , which will give rise to three new equations involving second derivatives of %, and the new functions
" (u) and
'' (v) .
In all we have now six equations from which,
in general, it is not possible to eliminate the six arbitrary functions. If such is the case, we must differentiate again, this time obtaining four new equations, involving third derivatives of z, and the two new * The student will note the difference between this definition and that given for a linear ordinary differential equation of the first order (§ 13) . See also § 87.
201
PARTIAL DIFFERENTIAL EQUATIONS
$ 77
"" (u) and y'" (v) . That is, we have now ten equations, from which the eight functions can be eliminated , in general, in two distinct ways . Hence we are led in the general case to two differential equations of the third order. functions
In the case of special forms of the function f, we can eliminate the six functions (u) , $ ' (u) , $ " ( u) , 4(v) , ¥' (v) , 4" (v) from the first six equations arising in the above process. Thus, for example, suppose f= w− [p(u) +4 (v) ] = o , where w is a known function of x, y, z. Then
στ dw + Əz p. дх
Ju
Ju
av
+ Əx
Əz
Ju მა Ju dw + q — $'(u) + 0+2) + 41 (2) ду მე ду
These last two equations involve only (u) and
(v) .
av + azP
av
=0,
მა
+ ду
' (u) and
9
= 0.
მი
(v) , and not
Hence, since the three equations gotten by differen
tiating these involve only p' (u) , p'' (u) , ¥ '(v), 4 " (v ) , these four functions can be eliminated from the five equations in which they enter and a single differential equation of the second order arises which has w = (u) + (v) for its primitive for any form of the functions and y.* Find the differential equations arising from the following primitives :
Ex. 1.
(x + y + %, x² + y² + z²) = 0 .
Letting x + y + z = u , x² + y² + z² = v, we have $(u, v) = 0 . * The five equations are linear in o ’ ( u) , o '' (u) , & ' (v) , ¥ '' (v) . Hence the elimination can be effected readily. Moreover r, s, t enter linearly also, and in such a way that the resulting differential equation is also linear in them , that is, it has the form Rr + Ss + Tt = V, where R, S, T, V, are functions of x, y, z, p, q.
202
DIFFERENTIAL EQUATIONS
$ 78
Differentiating, we get
аф № (2 x + 2 %p) = 0, Ju (1 + 1) + მა მი @ (1 + q) + მა (2y + 2 zq) = 0. du
•'. (1 + p) (y + zq) − ( 1 + 9) (x + zp) = 0,
or
(y  3)
+(
x) q = x  y.
Ex. 2. +(  xy, 2) = 0.
Ex. 3.
(x2 + y², zxy) = 0.
Ex. 4. 2
(x +y) +4(x − y) .
Ex. 5. z =
(x + y) +4(xy).
78. Solution of a Partial Differential Equation. Since a primitive which gives rise to a differential equation is obviously a solution of that equation, we see that arbitrary functions as well as arbitrary constants may enter into the solutions of partial differential equations. By the general existence theorem* for partial differential equations, it is seen that every partial differential equation , or system of such equations, has a solution containing a definite number of arbitrary functions. As an arbitrary function may contain an indefinite number of arbitrary constants, a solution involving an arbitrary function is much * This theorem is due to Cauchy, as is that for ordinary differential equations. Proofs of the theorem have also been given by Darboux and Mme . Sophie de Kowalewski, among others. The proof of the latter is the most readily followed and is the one usually given . See GoursatBourlet, Équations aux Dérivées Partielles du Premier Ordre, Chapter I ; also Picard, Traité d'Analyse, Vol . II , p . 318.
$ 78
PARTIAL DIFFERENTIAL EQUATIONS
203
more general than one that contains any fixed number of arbitrary constants. We shall speak of the solution given by the existence theorem (which is a solution involving one or more arbitrary functions) as the general solution.
The mere statement of the existence theorem for a system of partial differential equations of any order is quite complicated. We shall give here simply a statement in the cases of a single equation of the first and second orders in three variables : 1° Consider the equation f(x, y, z, p, q) = 0. We shall suppose that p actually appears.* Solve for it, so that the equation may be supposed to have the form p = F (x, y, z, 9).
The existence theorem tells us that if F (x, y, z, q) is regular † in the regions ofx = x0, y = yo, z = zo, q = 90, and if þ (y) is any arbitrarily chosen function ofy, regular in the region of yo, and such that (yo) = %0, ' (yo) = qo, there exists one and only one solution z = (x, y) , which is regular in the regions ofx = x0, y = yo, and which reduces to z = (y) for x = x0. Geometrically this means that given any curve z = (y) in the plane x = x0 there exists one and only one surface (in any region for which there are no singular points ofthe differential equation) passing through that curve. This can be generalized. By a proper choice of coördinates it can be shown that one integral surface, and only one, can be found passing through any arbitrarily chosen curve, whether plane or twisted (as long as we avoid singular points of the equation) . See Goursat Bourlet, p . 21 . 2º Consider the equation ƒ (x, y, z, p, q, r, s, t) = 0. If this contains neither r nor t, a linear transformation will introduce one or both of these. We shall
* Ifp is absent, then a must appear, and the argument here employed can be carried out by interchanging in it þ and 9, and x and y. The function F (x, y, z, q) is said to be regular in the regions of x = xo,y = yo 2 = 20, 9 = qo, if it can be developed by Taylor's theorem in a convergent power series in x  xo, y  yo, zzo, q 90. By a singular point of the equation we mean one whose coördinates together with the corresponding value of q determine a set of values in the regions of which F ceases to be regular.
204
DIFFERENTIAL EQUATIONS
$ 78
suppose then that one of them appears ; and there will be no loss in supposing that it is r. Solving for this, the equation takes the form
r = F (x, y, z, p, q, s, t) . The existence theorem tells us that if F is regular in the regions ofx = x0, y = yo, z = 20 , p = po, q = qo, s = so, t = to, and if p(y) and (y) are any arbitrarily chosen functions ofy, regular in the region ofyo, and such that p ( yo) = 20, ' ( yo) = 90, "' (yo) =·to, ¥( yo) = po, ¥' ( yo) = so, there exists one and only one function, z, ofx and y which is regular in the regions ofx = x0, y = yo, and such that z = (y) andp = 4 (y) for x = x0. Looked at geometrically, this means that given any curve z = (y) inthe plane x = xo, there exists an indefinite number of integral surfaces passing through it. But if at each point of this curve we fix a tangent plane, there exists one and only one integral surface through the curve and tangent to these planes. For, the direction cosines of the normal to the tangent plane are proportional to p, q,  I. As soon as the curve is given we know ( y) . The q at each point of this curve is p ' ( y) , hence it is also known. So that to give the tangent plane at each point is to give p, which is our ( y) . Once ø ( y) and ¥ (y) are given, the existence theorem says the integral surface is determined uniquely. As in the previous case, this may be extended to apply to any curve whether plane or twisted. Here again, no singular points of the differential equation are supposed to appear in the regions in which we are interested.
CHAPTER XIII PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 79. Linear Partial Differential Equations of the First Order. Method of Lagrange. — Lagrange deduced the following very neat method of solving linear partial differential equations of the first order. The general type of such an equation for one dependent and two independent variables is
Pp + Qq = R,
(1)
where P, Q, R are functions of x, y, z.
Consider the linear equation with three independent variables x, y, z,
(2)
P
Ju Ju du +R = 0, дх + e ду მთ
which is homogeneous (i.e. there is no righthand member) , and the coefficients are functions of the independent variables only.
If u = c satisfies ( 1 ) , u will be a solution of (2 ) ; for, differentiating,
alalals

Ju
we have
ди
ду du
Əz Substituting these in ( 1 ) we get ( 2) .
205
206
DIFFERENTIAL EQUATIONS
$ 79
Conversely, if u is a solution of ( 2 ) , uc will satisfy ( 1 ) .
For,
from (2) we have, on solving for R,
But
when u = c.
du
du
да P ди მე
αν du Q = R. az
ди ax du =p, Əz Hence ( 1 ) follows.
Ju
ὃν du Əz
9,
From this we see that the problem
of solving (1 ) is equivalent to that of solving (2) . Consider now the system of ordinary differential equations,
(3)
dx = dy P Q
dz R
If u is a solution of ( 2 ) , u = c satisfies (3 ) ; for, if we multiply du du du numerator and denominator of these fractions by respecdx' dy' dz tively, we have, by composition, that each of the fractions of (3 ) is equal to Ju dx du Ju + dy + dz дх az dy Ju Ju du P +R + Q дх Əz dy
Since the denominator is zero by hypothesis, the numerator (which is du) must be. Hence uc is a solution of ( 3 ) [ $ 65 , 3 °, ( b) ] . Conversely, if u = c is a solution of (3) , u will satisfy (2). differentiation we have
(4)
Ju ди du dx + dz = 0. dy + Əz дх ду
For, by
PARTIAL , OF THE FIRST ORDER
$ 79
207
To say that u = c.satisfies (3) is to say that for it
dx dy: dz
P : Q : R.
Hence, replacing in (4 ) the former by the latter, we have
Ju du pou + Q = 0, +R дх Əz ду which shows that ( 2 ) is satisfied . Hence the problem of solving (2) is reduced to that of solving (3). Therefore, finally, the problem of solving ( 1 ) is reduced to that of solving (3 ) , since every solution of the latter is also a solution of the former, and conversely. Moreover, if u = c₁, v = c, are any two independent solutions of (3), any function of u and ʊ, say ò̟ ( u , v) , will satisfy ( 2 ) .
For,
substituting this in the lefthand member of (2), we get
аф du
do P dv Ju Ju av Ju + +R P +Q + Q ·+ R. dv Əz дх ду дх Əz dy
This vanishes, since u and v each satisfy (2).
Hence
(u, v) is
also a solution of (2 ) irrespective of the choice of the function Therefore,
.
(u, v) = o * is the general solution of ( 1 ), since it con
tains an arbitrary function, § 78.
Since
is an arbitrary function,
there is no loss in putting zero for the righthand member instead of an arbitrary constant. Word for word, the above proof may be extended to a linear equation with n independent variables. So that we can formulate the rule : To find the general solution of Pr az მთ მ ax i+ P + ... + Pr Jx = R, дхг n d = x2= P2 Pi
dx solve
=
dxn Pr
dz =
R
* The solution may obviously also be written in the form u =¸=ƒ(v) , or v = 4(u)
DIFFERENTIAL EQUATIONS
208
$ 79
Ifthe general solution ofthis system is
== C1 , U₂ = C2, *** , Un = Cn then (U1, U2, … , u„ ) = 0 , where is an arbitrary function of U1, U2, ..., un, will be the general solution of(1).
Ex. 1. xzp + yz q = xy. dx_dy_dz = = XZ yz xy Multiplying numerator and denominator of the three members by y, x,  2 % respectively, we have, by composition, since the denominator vanishes, y dxx dy  2 z dz = o .
(Method 3°, (b) , § 65.)
.. xy — z² = C₁•
Besides,
dx dx dy ν = dy gives == == or C2. Xx XZ yz x y ☀ ( xy − z²,
Ex. 2.
y x
o is the general solution.
du Ju ди = 0. y + x + (1 + 2²) az дх dy dx = dy dz = du = x I + z² y
We have at once or
x dxy dy = 0,
x² + y² = €1.
Also,
u = C2.
PARTIAL , OF THE FIRST ORDER
§ 80
209
Multiplying numerator and denominator of the first two members by
y and x respectively, we have by composition,
x dy  y dx x² +y²
dz (Method 3°, (a), § 65.)
I +22 .. tan 12 x
tan1 % = C, y  xz
or
= Cz• x + yz u y², u, x² · `. * (~, ~ +5, 2 — 3 ) = 0, or × = ƒ(
+5 , x 2—32) + yz is the gen
eral solution . Ex. 3. yp  xq = x² — y².
Ex. 4. (y− 2) p + (z − x) q = x −y.
80. Integrating Factors of the Ordinary Differential Equation We are now in a position to treat satisfactorily
Mdx + Ndyo .
the problem of finding an integrating factor for an ordinary differential equation of the first order and degree .
We have seen ($ ( § 7)
that the necessary and sufficient condition that μ be an integrating factor of M dx + Ndy = o is Ə(µN) _ ¦(µM) = 0, дх ду
M or дх
(1)
ду
θμ M θμ + Noμ_ = 0, whence дх ду
θμ M N θμ = . ƏM ƏNƏx + ƏN ƏMay M * дх дх ду ду
* Since the number of solutions of this linear partial differential equation is infinite, we see again that an ordinary differential equation of the first order has an infinite number of integrating factors (§ 5) .
210
DIFFERENTIAL EQUATIONS
To find
§ 80
satisfying ( 1 ) we consider the system of ordinary
equations, ƏM ду
(2)
ON дх
ƏN дх dx =
N
ƏM ду αμ = ·dy = M μ
Remark. — In actual practice, when trying to integrate M dx + Ndy = 0 , we are not desirous of finding the most general form for µ ; as a matter of fact, as a rule, the simpler the form the better. Hence any one solution of ( 2 ) will be sufficient. It should be noted that this is not usually a practical method . But, for special forms of M and N, a solution of ( 2) can be found. Thus the following general classes of equations for which we can find a solution of ( 2) may be noted here. *
ƏM
ƏN дх 1° If ду is a function of x only, say f (x) , we have, from N f(x)da is an integrating factor (§ 17). (2), that μ ƏN ƏM дх ду 2° If π is a function of y only, say f (y), then μ μe M
is
obviously an integrating factor (§ 17) . 3° If the equation is linear, then M = Py  Q, N= 1 , and ( 2 ) du . P = eSPda is an integrating Hence με becomes Pdx = dy = = μ By Q factor (§ 13). 4° If M and N are homogeneous and of the same degree n, we get, by composition, after we have multiplied numerator and denominator of the first two members of ( 2 ) by y and x respectively,
ƏM ᎧᎳ y (yaM дх dy
dx +
x
ON дх
x
M Mdy dy
αμ
(3)
xM +yN
μ
* These were enumerated in Chapter II , in the list of equations for which integrating factors can be found.
PARTIAL, OF THE FIRST ORDER
§ 81
211
By Euler's theorem for homogeneous functions, we have x
ƏM OM = nM, +y дх ду
..y
ᎧᎳ IN == nN, +y дх ду
... x
x
JM
= nM  x
ду
ON дх
әм дх
IN = nN  yду
whence (3 ) may be written
ƏM ᎧᎳ әм IN dx + dy) ++ 1 ( Jx dx + ―n (Mdx + Ndy) дх дх ду dy) ду
x
x M + yN
αμ μ
(Mdx + Ndy) But Mdx + Ndyo. Hence we may add (n + 1 ) (M to the numerator on the left without altering its value. Doing so, we have d(xM +yN) = ― αμ . xM +yN I
Integrating, we have
μ=
(§ 17) . xM +yN
5° If M = yfi(xy) , N = xf2(xy) , then on multiplying numerator and denominator of the first and second members of ( 2) by y and x respectively, we have by composition , after obvious reductions,
_dp_ d(xM  yN) = xM  yN μ I
On integrating, we have
(§ 17) . ·
μ=
xM  yN
81. Non linear Partial Differential Equations Complete, General, Singular Solutions. a primitive
(1)
(x, y, z, a, b) = 0
of First Order.
We have seen (§ 76) that
212
DIFFERENTIAL EQUATIONS
$ 81
which involves two arbitrary constants gives rise to a unique partial differential equation of the first order
(2)
f(x, y, z, p, q) = 0.
This differential equation is gotten by eliminating a and b from (1) and its derivatives with respect to x and y respectively ; viz.,
(3)
aф d +1 Э +p дх аф +9 ду
аaфz = 0, 26
= 0. Əz
(1 ) is spoken of as a solution of ( 2 ) . constants.
Here , of course, a and b are
Letting them be parameters, we have, on differentiating
( 1) , and taking account of (3) ,
дф да афаз = 0, + ab ax da dx (4)
афаз дф да = 0. + да ду ab ay
These two equations can be consistent only in case either
Эф = 0, да (5)
gle els
მი = 0, ab
да ax
ab дх
da
ab
ду
ду
or
= 0.
PARTIAL, OF THE FIRST ORDER
$ 81
213
If this determinant vanishes, b is some function of a, say ¥(a), (Note I in the Appendix) .
(6)
Then (4) may be replaced by b = 4(a), Эф аф +4'(a) = 0. да მხ
Since (5) and (6) were gotten on the assumption that (3 ) hold, it follows that if we eliminate a and b from ( 1 ) and (5) , or from ( 1 ) and (6), we shall get relations between x, y, z which will satisfy (2) . Hence these relations are also solutions. We see then that the primitive of a partial differential equation of the first order is not the only solution .
But since the others can be
gotten from it, Lagrange called it the complete solution. We have already noted ( § 78) that, in the general theory of partial differential equations, it is proved that an arbitrary function appears in the general solution of an equation of the first order.
Since in (6)
the function (a) is any function of a we please, Lagrange called the solution gotten by eliminating a and b from (1 ) and (6) the general solution. A particular solution is gotten by assigning definite values to a and
in the complete solution, or by using a definite function
(a)
and eliminating a and b from ( 1 ) and (6) . On the other hand, the solution gotten by eliminating a and b from (1 ) and (5) contains nothing arbitrary, and is known as the singular solution.
It is the exact analogue of the singular solution
of ordinary differential equations.
Looked upon geometrically, it is
the equation of the envelope of the doubly infinite number of surfaces whose equation is given by (1) , just as the general solution is the • envelope of an arbitrarily chosen single infinity of those surfaces. It can also be shown * that the singular solution can be gotten from the differential equation by eliminating p and q from af af =0, = 0, f(x, y, z, p, q) = 0, aq ap * See GoursatBourlet, p . 24, also p. 199 and foll.
DIFFERENTIAL EQUATIONS
214
$81
in exact analogy to the case of ordinary differential equations of the first order. But the limits of this book prohibit a further discussion of the subject. remarks :
We shall conclude it with the following obvious
1° There need be no singular solution.
This will happen in case
the equations (1 ) and ( 5 ) are inconsistent ; or, geometrically, in case the surfaces (1 ) have no envelope. 2º The general solution cannot be written down.
(a ) is entirely
at our disposal ; but until it is given, there is no way of eliminating a and b from (1) and (6), and when it is given, we have, of course, a particular solution . 3° There is no unique complete solution.
Any solution involving
two arbitrary constants may be taken as one.
It is easy to see that
there is an indefinite number of them . for
For, if we choose any form
(a ) which involves two arbitrary constants h and k, on eliminat
ing a and b from ( 1 ) and ( 6 ) we get a solution involving these two arbitrary constants, which fulfills all the requirements for a complete solution.
We saw in (§ 76) that a complete solution of 22 (p² + q² + 1 ) = R² is
 a)² + ( − b) ² + 2² = R², (x − which represents a family of spheres of radius R and centers in the plane z = 0. The envelope of these is the pair of planes z² = R², which is obviously the result of eliminating a and b from (x  a )2 + ( y  3) 2 + 2 = R, x  a = 0, y  b= 0, or of eliminating p and q from ≈² ( p² + q² + 1 ) = R², z²p = 0, z²q = 0.
22 = R2 is then a singular solution. The choice of any relation b = (a ) results in selecting those spheres whose centers lie along the curve y = (x) in the plane zo. The envelope of these is obviously the tubular surface traced out by the motion of a sphere of radius R
$ 82
PARTIAL , OF THE FIRST ORDER
215
moving with its center on the curve y = (x) and along its entire length. In particular, if (x) is a linear function of x, the envelope is a cylinder. As an example, suppose b = ha + k. To find the corresponding solution, we have to eliminate a from (xa)² + (y  ha  k)2 + 22² = R², (xa) + hy  hak) = 0. From the second of these we have a = x + ky  hk I + h² Putting this in the other equation, we have
h2 (hxy + k) ² + (hx − y + k) ² .= R2 — 22, (1 + h²)2
or
(hx − y + k)² = ( 1 + h²) (R² − z²) ,
which is obviously the equation of a circular cylinder whose axis is the line y = hx + k, z = 0. Since this solution contains two arbitrary constants, it is a complete solution. (See Ex. 6, § 83.) As an exercise, the student should start with (hx − y + k)² = ( 1 + h²) ( R² − z²) as a primitive and show that, considering hand k as the arbitrary constants, the resulting partial differential equation is again z² ( p² + q² + 1 ) = R².
82.
Method of Lagrange and Charpit.  Since the knowledge of a
complete solution is sufficient to enable us to find all other solutions, it is usually desirable, whenever possible, to find it first. suggested the following method : Given the equation (1)
f(x, y, z, p, q) = 0,
Lagrange
216
DIFFERENTIAL EQUATIONS
§ 82
try to find a second relation p(x, y, z, p, q, a) = 0 ,
(2)
which involves either por q or both, and also an arbitrary constant, ? and put these and such that when we solve ( 1 ) and ( 2 ) for p and q in the total differential equation
dz = p dx + q dy
(3)
the latter can be solved ; the solution of ( 3 ) is a complete solution of (1 ) , since, from what precedes, it determines z as such a function of x and y, that the values of p and q obtained from it, together with z satisfy (1) ; besides it involves two arbitrary constants. we may proceed as follows :
To do this
Differentiating ( 1 ) and (2 ) with respect to x, we have
(4)
(5)
af af af dp af dq = 0, * + +p მთ + дх dp dx dq dx Эф d аф афа p dр p , do dq = 0. + + +p Əz дх op dx oq dx
Similarly, differentiating with respect to y,
(6)
af af dp + af dq = af +9 az + ap dy dq dy 0, дуаф
+9
(7)
ay
Эф do dp + az ap dy
do dq = 0. aq dy
Now the condition that (3 ) shall be integrable is obviously dp dy
(8)
༤8
* Throughout this section, dx
dq = 0. dx
a +p a" d Эх Əz dy
a dy
a• Əz
PARTIAL, OF THE FIRST ORDER
§ 82
217
From equations ( 4 ) , ( 5) , (6) , ( 7 ) , (8) we can eliminate the four quantities dp, dp, dq, dq . We get rid of dp by multiplying (4) dx dx' dy' dx' dy Эф and af and (5 ) by respectively, and subtracting. The result is ap ap
af +p алаф . 0%) Әр ax
(9)
(дх
Multiplying (6) and (7) by
0% ) Әр
+ ´aƒ a& aq ap
a& aƒ\dq = 0. aq opdx
მა and af respectively, and subtractƏq
aq
ing, we get rid of dq and have dy
(10) (01/1
+9
afab  მდ a&\ af +9 + ´aƒ a& az aq ду მუ მ q მტ მი
a aƒdp = 0. op dq dy
Making use of (8) , we have, on adding and arranging according to derivatives of p,
(11)
af +p +9 af ad ax ду az ap + ( მ≈  მი 噐)
af aaƒ a др дх да ду
afa& = 0. ·+ 9 მემო ( 00+
This is a linear equation.
Hence, to solve for 4, we consider the
system of ordinary equations
( 12)
dp
dx
dq = of
af
dy af
af af af+9 Эх მე Á მი ду Әр ‡ ' Á ‡
dz
do
ap ** ¿
DIFFERENTIAL EQUATIONS
218
§ 82
Our aim is to find, not the most general form of 4, * but any form of it that contains p or 9 , and an arbitrary constant. Hence, in actual practice, we look for the simplest one. Remark.
It is desirable that ( 11 ) or ( 12 ) be committed to memory.
Ifwe put
d= a a and d = a +9 , +1. dx Эх ဝါး dz dy dy
we have the skew symmetrical form
dx (12').
af ap
= dq df dy

df dx
dy af მი
dz as for the next term
this is really a result of the equality of af a + f ap 9 Jq
dx
dy as may be seen by composition after multiplying numerator and af  af მი ap denominator of the first by p and of the second by 9, and taking account of (3) . Or using the same notation as before, and writing df op dx ap
af do = [f, ap dx
]z, p
df de  af do == [ƒ, $ ]y, q dy dq aq dy
(11 ) may be put in the compact form (11')
[ƒ, $ ]x, p + [ ƒ, $ ]y, q = 0. ·
* Lagrange thought originally that by using the general form of (which would involve an arbitrary function) he could get the general solution by this method. But this is, in fact, not practicable. Charpit, in a memoir presented to the Académie des Sciences, in June, 1784, suggested the use of any form for involving p or q and an arbitrary constant, obtaining by its use the complete instead of the general solution.
$ 82 Ex. 1 To find
PARTIAL, OF THE FIRST ORDER
219
zpq= 0. we must find a solution of
dp_ = dq_ =dx  dy P 9 9 P
Using the first two members, we have
= 0. $ = p − aq =
Combining this with the original equation , we have
ช 9q= V a
p=a
L& T8
Then (3) becomes dz = a
a dx + √ dy,
Va dz = a dx + dy.
or Z Integrating, we have
2 Vaz = ax + y + b,
or
4 az = (ax + y + b)²,
which is a complete solution. If we had used the second and third members, we would have had
$= q− x α = o.
q = x + a, and p = Then (3) becomes dz 
Whence
ช x + a
2. dx + (x + a) dy, x + a
or
dz z dx dy = (x + a ) (x + a) ²
DIFFERENTIAL EQUATIONS
220
§ 82
Z Integrating, we have y + b ==
x + a or
z = (x + a) ( y + b),
which is also a complete solution. The general and singular solutions can be gotten from either of these. Thus, using the second one, we get the general solution by eliminating a from z = (x + a) [ y +4(a) ]
and where
y +4(a) +4' (a) (x + a) = 0, (a) is any function of a.
In particular let the student show that if we put
4(a) = k − ha where h and k are any constants, the corresponding solution is 4 hz = (hx + y + k)², which we recognize as the first form obtained.
The singular solution , resulting from eliminating a and b from
z = (x + a)(y + b) , y + b = 0,
x + a = 0,
is z = 0. Let the student show that this can also be gotten from the other form of solution.
Ex. 2.
 ( + yq)².
Ex. 3. √p + √q = 2 x.
883
PARTIAL, OF THE FIRST ORDER
83. Special Methods.
22I
Special methods for certain forms of the
differential equation at times prove simpler than the general method of § 82 (although most of them are suggested by the latter) . Some of these are the following : 1° Suppose all the variables absent.
The equation takes the form
f(p, q) = 0,
(1)
and the equations for determining & become
dp_dq = O
(2)
From the first member we see that pa. Then is gotten by substituting in (1 ) . bis determined byf(a, b) = 0. We have then
Evidently q = b, where
dz = adx + bdy, z = ax + by + c, where ƒ (a, b) := 0.
or Hence the rule :
The complete solution off(p, q) = o is z = ax + by + c, where ƒ ( a, b) = 0.* · By means of simple transformations, certain forms of equations can be brought into this type : dz =e202 = 2 ᎧᏃ ; so that Putting log z = Z, or z = e², we have p = Эх дх дх
L_ = Əz 2 дх Similarly
4 = ᎧᏃ . 2 ay
* The complete solution represents a doubly infinite set of planes . Any particular solution represents the envelope of a chosen single infinity of these. This is a developable surface. There is no singular solution . Why ?
222
DIFFERENTIAL EQUATIONS
$ 83
= 0, log z = Z will transform it into Hence, if the equation is f ƒ (2,2 )= az az 1(32,82)= a
Again, if we let log x = X, or x = e³,
p =
Əz = dz dx  I dz . i.e. xp = 3s .. ax дх Jx dx xox
Similarly, putting log y = Y, we get yq=
Əz ay
Hence
g = X; = by lo x = ) =0
X' f(xp, q) = 0 is transformed into f (Ox Əz f(p, yq) := o is transformed into f O
Əz
= 0 by log y = Y; az = 0 by logx = X, log y = Y;
ƒ(xp, yq) = 0 is transformed intoƒ(3
az o = 2; 32 ) = 0 by log x = X, log == ƒ ( 32 , 22) f(xp, 2) = 0 is transformed into/(32, and so on.
Ex. 1. pq = I. The complete solution is z = ax + by + c, where ab = 1 ,
I y + c. z = ax + 1/2 y
or
Ex. 2. q =% + px. x,9 we see that the substituWriting this in the form 2 = 1 + px ช ช tion X = log x, Z = log z, will transform this equation into
ᎧᏃ
=1+ ду
ᎧᏃ ах
§ 83
PARTIAL, OF THE FIRST ORDER
223
The complete solution is Z = aX + ( 1 + a) y + c. Passing back to the original variables, we have log za log x + ( 1 + a) y + c, or z = bxae(1+a)y.
Ex. 3. px + qy = 1. Ex. 4. x²p² + y²q² = z².
Ex . 5. yq = p. 2° If x and y are absent, the equation takes the form
f(z, p, q) = 0,
(1)
and we have for determining
dp = dq = af af P Əz 9 მე
(2)
dp= _dq, whence q = ap. " 9 Substituting in ( 1 ) , we have ƒ(z, p, ap) = o, whence p = 4(z, a), dz dx + a dy, and dz = p dx + q dy becomes 4(z, a) where the variables are separated . To put this rule in shape easily to be carried in mind , we note that, to say q = ap is to say that z is a function of x + ay, by the general method of § 79. If we put x + ay = t, we have
p=
მშ дх
dz dt
= Əz = a dz dt' ду
and he equation ( 1 ) becomes the ordinary differential equation
dz dz a = 0, f %, dt' dt
DIFFERENTIAL EQUATIONS
224
$ 83
Hence the variables dz We have, then, can be separated immediately after solving for dt the rule : in which the independent variable is absent.
If the equation is ofthe form f(z, p, q) = 0 , put x + ay = t, which dz dz will replace p by dt " and q by a dt Put these values in the equation dz and solve for dt If the equation has one of the forms f(%, xp, q) = 0, ƒ(%, p, yq) = 0, ƒ (z, xp, yq) = 0, one, or both of the substitutions log x = X, log y = Y will reduce it to the above form .
Ex. 6. z²(p² + q² + 1 ) = R². dz dz and the equation becomes Putting x + ay = t, p = dt + ' 9 = a ;dt 22
2 (1 + a²)) + 11 = R³, ] VI + a²zdz == dt.
or
VR222 ..  √i + a² √R²  2² = t + b = x + ay + b,
or
(1 + a²) (R² −z²) = (x + ay + b)².
Ex. 7. xp(1 + 9) = q%.
Ex. 8. z = pq. 3° If the dependent variable is absent, and the equation is such that it can be put in the form
(1)
f(x, p) =f2(y, 9)
§ 83
PARTIAL, OF THE FIRST ORDER
225
(a sort of separation of the variables) , the equations to determine become dp = ... = dx = (2) of of дх ар
əfi dx + afi dp = o, or df = o. дх ap
Hence we have
f₁ = a. .. f₂= a. Solving these, we have p and dz = p dx + q dy becomes
(x, a) , q = 4½(y, a),
dz = 4₁(x, a)dx +42(y, a)dy, in which the variables are separated.
Hence the rule :
If the dependent variable is absent, and the other variables are separable, such that the equation takes the form f₁ (x, p) =ƒ½ (y, 9) , equate each of these members to a constant, solve the resulting equations forp and q, and put these values in dz = p dx + q dy.
the transformaIf the equation can be put in the form fi ( x, 2)= tion log z = Z will reduce it to the form above.
(y, 2),
Ex. 9. q = 2yp². Ex. 10. 2 (2x  zy) —p + q = 0. 4° The equation z = px + qy +ƒ(p, q) , which is usually referred to as the extended Clairaut equation ( § 27) , will obviously be solved if we put p = a, q = b. We have then the complete solution z = ax +by +f(a, b).
226
DIFFERENTIAL EQUATIONS
§83
While the general method of § 82 applies here, it does not give this simple form of solution. By that method we may use either pa or q = b, but not both simultaneously. As a matter of fact it is an accident if the result of substituting in the differential equation the values of p and q obtained from two solutions of equations ( 12 ) , § 82 is a complete solution. It does happen, at times, as in the case in question. But there is no certainty that it will, nor is there even a likelihood of it.
Ex. 11. Solve z = px + qy + √p² + q² + 1 , and examine for singular solution .
5° If the equation is of the form ƒ(x + y, k, q) = 0 , let q = p + a. Thenf(x +y, p, p + a) = o gives p = 4(x + y, a), whence q= 4(x +y, a) + a, and the equation dz = pdx + q dy becomes dz = 4(x + y, a) (dx + dy) + a dy. Let the student show that this form of solution is given by the general method of § 82. If the equation is of the form f (x +3 = 0, the transformation log z = 2 f( x + y, 2, Z 2) will reduce it to the form above.
Ex. 12. p(x +y) q = 0. Ex. 13. zp(x + y) + p(q − p) = z². 6° If either
or q is absent, the method of solution is obvious
by inspection.
In the former case integrate considering x as a con
stant, when the constant of integration will be an arbitrary function of x. In the other case integrate considering y as a constant, when the constant of integration will be an arbitrary function of y. These solutions, involving arbitrary functions, are general solutions. * * When these equations are of the first degree in the derivative, they are linear equations. The method here given is exactly that of Lagrange for such equations ($ 79) .
§ 84
PARTIAL, OF THE FIRST ORDER
227
Ex. 14. (xy)q − (x + z) = 0.
dz Considering x as a constant, we can write q = dy dz
dy
x+ 2
x y
.*.
= 0.
Integrating and taking exponentials of both sides, we have (x + z)(x − y) = $(x), where p(x) is an arbitrary function of x.
Ex. 15. xp Ex. 16.
2 zp + xy = 0. (z  x) = 0 .
Ex. 17. y²(p² — 1 ) = x² p².
84. Summary. Partial differential equations of the first order are divided into two general classes : those which are linear in the derivatives of the dependent variable, and those which are not. 1° For the solution of linear differential equations of the first order the method of Lagrange applies, giving the general solution ($ 79). 2° For the solution of nonlinear equations of the first order, the general method of Lagrange and Charpit applies ($ 82 ) , giving a From this the other solutions can be gotten complete solution. (§ 81 ) . At times the special methods of § 83 are shorter than the general method of § 82. Sometimes a transformation of variables will help in the solution of an equation.
DIFFERENTIAL EQUATIONS
228 Ex.
1.
Ex.
2. q = p² + 1.
Ex.
3. x
$ 84
x² + y²q = z² .
du ди Ju +y +% = xyz. дх ду მ
Ex. 4. z = px + qy + (p + q)².
Ex.
5. xypq = 22.
Ex.
6. y²zp + x²zq = y²x.
Ex.
7. q = xp +p².
Ex.
8.
Ex.
9. (x + y)(p + q)² + (x − y) (p − q)² = 1 .
( +9) (px + qy) = 1. [Let x +y = u²,
x − y = v² ]
Ex. 10. (
+ 9² ) x− pz = 0.
Ex. 11. (x² + y²) (p² + q²) = 1 .
[Let x = p cos 0, y = p sin 0. ]
Ex. 12. ( y² + z² − x²)p − 2 xyq + 2 xz = 0.  q) . = (p − Ex. 13. q²²
Ex. 14.
(y  x) (qy — px) = (p − q) .
[Let xy = u, x + y = v.]
Ex. 15. z  xp  yq = 2 √x² + y² +z².
Ex. 16. pq = px + qy. du Ex. 17.
(y + z + u)
Ju + (x + u + x)
дх
ду
du + (u + x + y) : Əz = x+y+%
$ 84
PARTIAL, OF THE FIRST ORDER
229
Ex. 18. Determine a system of surfaces such that the normal at each point makes a constant angle with the plane of xy. Ex. 19. Determine a system of surfaces such that the coördinates of the point where the normal meets the plane of xy are proportional to the corresponding coördinates of the point on the surface. Ex. 20. Determine a system of surfaces for which the product of the distances of the tangent plane from two fixed points is a constant.
CHAPTER XIV PARTIAL DIFFERENTIAL EQUATIONS OF HIGHER ORDER THAN THE FIRST 85. Partial Differential Equations of the Second Order, Linear in the Second Derivatives . Monge's Method. — The general type of a partial differential equation of the second order linear in the second derivatives is Rr + Ss + Tt = V, (1) where R, S, T, V are functions of x, y, z, p, q. Gaspard Monge ( 17461818) suggested a method, which is known by his name, by which a first or intermediary integral is found in the form of a partial differential equation of the first order involving an arbitrary function. The solution of this equation by any of the methods of Chapter XIII , or otherwise, will then give the general solution. While this method applies only in case R, S, T, V satisfy certain conditions, it works sufficiently frequently to justify our giving here at least the rule by which solutions are gotten by this method . * Besides (2)
dz = p dx + q dy,
we have
= { db = rdx + sdy, s dx + t dy.
(3)
Eliminating
and
from ( 1 ) and (3 ) we have
(4) s(R dy² — S dx dy + T dx²) − (R dy dp + T dx dq — V dx dy) = 0 . * For a detailed account of this subject see Forsyth, p. 358 and foll., or Boole, Chapter XV. 230
$ 85
PARTIAL, HIGHER ORDER THAN THE FIRST
231
Whenever it is possible to satisfy simultaneously, (5)
Rdy2  S dx dy + Tdx² = 0,
(6)
R dy dp + Tdx dq Vdx dy = 0, *
(4) will be satisfied and, therefore, so will ( 1 ) .
(5 ) is equivalent to
two equations of the first order, (7)
dy — W₁ (x, y, z, p, q) dx = 0 , dy — W₂(x, y, z, p, q) dx = 0,
which become identical in case (8)
4 RT= S².
Equations ( 2 ) and ( 6 ), together with either one of ( 7 ) , constitute a system of three total differential equations in the five variables x, y, %, p, q.
Such a system can be solved only in case certain conditions
are fulfilled, and it is for this reason that Monge's method does not always work. It will work if we can find two independent solutions of this system u₁(x, y, z, p, q) = C₁,
u (x, y, z, P, Q) = Cg.
In this case it turns out that (9)
u₁ =
(U₂),
where is an arbitrary function, is an intermediary (or intermediate) integral. Looked upon as a partial differential equation of the first order, ( 9 ) must be integrated again . solution of (1) .
Its general solution will be the
In case it happens that not only one of (7) , but each one, together with ( 2) and (6) , determines a system that can be solved, we have * Equations (5) and (6) are usually referred to as Monge's equations.
DIFFERENTIAL EQUATIONS
232
two intermediary integrals (9 ) .
$ 85
Solving these for p and 9, we put
the values of the latter in dz =p dx + q dy.
The integral of this is
the general solution of ( 1 ) .
Ex. 1. qr  2 pqs + p³t = 0. Monge's equations are q² dy² + 2 pq dx dy +p² dx² = 0, q dy dp +p² dx dq = 0.
The first of these is a perfect square , (q dy + p dx)² = 0 . Substituting this in the second one, it becomes
qdp  p dq= 0,
=
whence
9 The first one, combined with dz = p dx + q dy, gives
dz = 0, whence
% = C2.
Hence an intermediary integral is p = qp(z). In this case we have only one intermediary integral, hence we must integrate this. Since it is linear, it can be solved by the method of Lagrange (§ 79) .
Its general solution is
xp(z) + y = 4(%).
*
$ 85
PARTIAL, HIGHER ORDER THAN THE FIRST
233
Ex. 2. r a²t = 0. * Monge's equations are
dy²  a² dx²:= o, or dy — a dx = 0 and dy + a dx = 0, dydp — a² dx dq = 0 . Using dy  a dx = o, we have y  ax = c1.
Combining this with the second of Monge's equations, we get dpa dq = 0 ; whence p  aq = Cq. Hence an intermediary integral is p― aq = 4(y — ax) .
Using the other equation , dy + a dx = o, we get a second intermediary integral
p + aq = p(y + ax).
Solving these for p and q, we have
I p == [p (y + ax) +4 (y— ax)], I • = 2 [ $ (y + ax) − 4 (y — ax)] . A much simpler method of solution for this equation will be given in § 88. This equation plays an important rôle in Mathematical Physics. It was first integrated by JeanleRond D'Alembert ( 17171783) in a memoir entitled Récherches sur les vibrations des cordes sonores, presented in 1747 to the Berlin Academy. In studying the vibrad2y a²y = 0, tions of a stretched elastic string, he considered the equation in the form dt2 dx² where t is the time, and x and y are the rectangular coördinates of a point of the string, x the coördinate measured along the line joining the extremities of the string, and y the displacement of the point from the position of equilibrium . His proof is given in Marie, Histoire des Sciences Mathématiques et Physiques, t. VIII . p. 217.
DIFFERENTIAL EQUATIONS
234
$ 85
We have now to solve I − 4(y — ax) ] dy. dz = = [p(y + ax) +4(y − ax)]dx + 2—a [ $ (y + ax) — I == $ (y + ax)(dy + adx) +4 (y − ax ) (dy — a dx), which is exact.
Since
and
are symbols of arbitrary functions, we shall retain
them in writing the solution z = $(y + ax) +4(y — ax). There is no loss in failing to add an arbitrary constant, since either of the arbitrary functions may be supposed to incorporate that.
Ex. 3. r t= 
AP . x +y
Monge's equations are dy² dx² = o, or dy — dxo and dy + dx = 0, dydp
 dx dq + 4Pdx dy = 0. x +y
Using dy  dx = o, we have y  x = c₁.
Combining this with the second of Monge's equations, we have 2 x dp +4 pdx − 2 x dq + c₁ (dp — dq) = 0.
Also
dz = p dx + q dy becomes dz = p dx + q dx.
Subtracting twice this from the above equation, we get 2 (x dp +p dx) 2 (x dq + q dx) + c₁ (dp− dq) + 2 dz = 0.
$ 85
PARTIAL, HIGHER ORDER THAN THE FIRST
235
This is exact, and has for solution (2 x + c₁) (p − q) + 2 % = C₂, or
(x + y) (p− q) + 2 % = C₂• Hence an intermediary integral is (x + y)(p − q) + 2 z =
(y  x).
Using the equation dy + dx = o, we get a system of total differential equations which are not integrable.
Hence we must integrate
the intermediary integral. This is linear, so Lagrange's method applies, dx  dz  dy = (y x) 2z x+ y x+y
From the equation of the first two members we have x + y = a. Replacing y by its value a  x, we have
dx = α (a or
dz 2x)  2 %'
dz 2 I + ² % = = p(a − 2 x) . dx a a
This is a linear ordinary equation of the first order. 2x ting factor (§ 13) is eafdrea , and the solution is
An integra
2x az ea = =Sea 4 (a − 2 x) dx + b. Replacing a by its value x + y, we have the general solution 2x (x + y)zex+y Sea $(a− 2 x)dx = 4(x + y).
DIFFERENTIAL EQUATIONS
236
$ 86
Here, as in the case of the nonlinear partial differential equations of the first order ( § 81 ) , the general solution cannot be written down. For until the form of is known the above integral cannot be calculated.
In any example in which the initial conditions de
termine 4, the a which appears in the integral must be replaced by x + y, after the integration has been effected.
Ex. 4. q( 1 + q) r − (p + q + 2 pq)s +p(1 +p)t = 0.
Ex. 5. ps  qr = o. Ex. 6. (b + cq)³r − 2 (b + cq) (a + cp) s + (a + cp)²t= 0.
86. Special Method .
At times, by considering one or the other
of the independent variables as a constant temporarily, the equation may be looked upon as an ordinary differential equation.
Of course,
an arbitrary function of the variable supposed constant must take the place of the arbitrary constant in the solution. examples will illustrate :
The following
Ex. 1. xr = p• Letting y be a constant temporarily, this may be written
x x dp =p, or dp_d dx Д x Integrating, we have pxf(y) , where ƒ(y) is an arbitrary function. Again letting y be constant, we have dz :xf(y), dx whence z = x²f(y) + Here the factor
(y), where
(y) is another arbitrary function.
, arising on the right, is incorporated inƒ(y) .
$86
PARTIAL , HIGHER ORDER THAN THE FIRST
237
Ex. 2. r + s + p = 0 . Integrating this, considering y as a constant, we have
p + q + z = ƒ(y). This is linear and of the first order.
Hence Lagrange's method
applies.
dx I
dy I
dz f(y) — %
From the first two members we get x ― y = a. From the last two members we have the linear ordinary differential equation
dz + z = ƒ(y). dy An integrating factor is e (§ 13) , and the solution is
z =S/(y)edy + b = $(y) + b. Hence the general solution is
ze³ — $(y) = 4(x − y), or
z = $(y) + e¯³4(x − y),
where the factor e is incorporated in
Ex. 3. yt  q = xy².
Ex. 4. Sxy. Ex. 5. r + p = xy.
(y).
DIFFERENTIAL EQUATIONS
238
§ 87
87. General Linear Partial Differential Equations.  We shall consider now partial differential equations which are linear in the dependent variable and all of its derivatives. The general type of such equations is
Jnz
Jnz (1)
+ P 1, 1
n, 0
dxn
+ Pn1,0
anz
anz + Pn 2, 2
მე 1 მ y
+
+ Po, n
Əxn2 Əy²
Əyn
ar+sz Jn 1z მო + ... + Ps, + ... + P1,0 8, r dxn1 дх მ მ y" მთ + Po, 1
+ Po, 0 % = f(x, y),
ду where the coefficients are functions of x and y, including the case where some or all of them are constants. If we put D=
a дх
= a (1) may be written ду
(Pn, 0D" + Pn1,1D"¹D + Pn  2, 2Dm ² D² + ... + Po, nd" + P₂  1, 0Dn−1 + ... ··· + P„‚ ‚DºD” + ... + P1,0D + Po, 1d + Po, 0) z = f (x, y),
or more briefly
(1) where F(D,
F(D, D) z =f(x, y), ) is a symbolic operator, which, looked upon algebra
ically, is a polynomial of degree n in D and D.
There are many
points of similarity between this equation and the linear ordinary differential equation of the nth order (§ 42).
Obviously,
F(D, D) (u + v) = F (D, D) u + F(D, D) v.
§ 88
PARTIAL , HIGHER ORDER THAN THE FIRST
239
Hence the problem of solving ( 1 ) can be divided into two, viz . that of finding the general integral of
F(D, D) % = 0,
(2)
which we shall call the complementary function of ( 1 ) , and that of finding any particular integral. eral integral of ( 1) .
The sum of these will give the gen
88. Homogeneous Linear Equations with Constant Coefficients. ― Following a generally adopted convention, we shall use the term homogeneous to apply to an equation in which all the derivatives are of the same order.
In this case the symbolic operator is homoge
neous in D and D. Suppose, besides, that the coefficients are constants, and the righthand member zero. Our equation will be of the form
(1)
(k。D™ + k₂D”−¹D + ... ··· + kn  1DD”¹ + k„D”) z = 0,
or
F(D, D) % = 0.
Since for
any function whatever DrD³p (y + mx) = m”$(r+ ) (y + mx), dr ++s (y + mx) , the result of substitut[d(y + mx)]r+s³
where
+ (y + mx) means
ing z =
(y + mx) in (1 ) will be $(n)(y + mx)F(m, 1) = 0.
Hence z =
(2)
(y + mx) will be a solution, provided F( m, 1 ) = 0 ; i.e. k。m" + k₁m²1 + ... ··· + kn1m k₂ 1m + kn k₂ = = 0.
DIFFERENTIAL EQUATIONS
240
§ 89
If the roots of (2) , which we shall speak of as the auxiliary equation, are distinct, say m₁, m2, " Mn z = 41(y + m₁x) + $₂(y + m¸x) + ... ··· + $n (y + m„x) will be a solution. Since it contains n arbitrary functions , it will be the general solution. *
Thus let us consider the equation in Ex . 2 , § 85, 22% dx²
a2
22% = 0. ay²
The auxiliary equation in this case is
m² — a² = 0.
... m = ± a.
Hence the general solution is z = $(y + ax) +4(y − ax).
Ex. 1.
Ex: 2.
22%
22%
02% J.x2
3
23% მე 3
7
= 0.
+2
Әх ду
მე
ეპი +10
Əx² dy
J3% = 0. ах дуг
89. Roots of Auxiliary Equation Repeated.
If any of the roots
of the auxiliary equation are repeated , the method of § 88 fails to give us the general solution . In this case we proceed by a method entirely analogous to that in § 47. as a factor, F (m, 1 ) is only of degree n— I. The lost * If F(D, ) contains root in this case is ∞ , and the corresponding integral is (x) . This is also obvious is a factor of from the form of the differential equation in this case. For to say that F(D, D) is to say that every derivative of ≈ is taken at least once with respect to y. Hence z = (x) will give o. Similarly, if F (D, D) contains Dras a factor, 1 (x), yo2(x), ***, yrlor (x) are readily seen to be integrals.
$ 89
PARTIAL , HIGHER ORDER THAN THE FIRST
241
The symbolic operator F ( D, λ) may be written as the product of its factors (D — m²D) (D — M₂D) ... ••• (D — m„D) . Moreover, it is readily seen (§ 46) that the order of these factors is immaterial . Suppose m₁ a repeated root. We wish to find a solution of (D — m₁ð) (D — m₁D)z = 0. Putting (D  m
) z = v, our equation is
(D — m₁D)v = 0. Hence, by the method of § 88 We now have to solve
v = $(y + m₁x).
(D— m₁D)z = $(y + m₂x).
This is linear and of the first order, pm₁q =
(y + m₁x).
Hence the method of Lagrange ($ 79) applies, dx I
dz dy = mi (y + m₁x)
From the first two members we have y + m₁x = a. Putting this in the last member, we have
dx =
dz • $(a)
..xp(a) — z = b. Hence the general solution is xp(y + m,x) − z = 4(y + m₁x) ,
or
z = 4( y + m₁x) + x$(y + m₁x).
DIFFERENTIAL EQUATIONS
242
$ 90
In other words, if m, is a double root of the auxiliary equation, not only is
(y + mx) an integral, but so also is x4( y + m₁x) .
In an
entirely analogous manner it can be shown that if m₁ is an rfold root, $1(y + m₁x), xp2(y + m₁x) , x²Þ3( y + m₁x) , ... , x²¹µ,(y + m₁x) are all integrals.
Ex. 1.
22% dx2
22% 2
22%
J3% + dxdxdy ду J3%
Ex. 2.
Ex. 3.
= 0.
+ дх ду
дуг 23%
23% = 0.
Ахдуг dxdy²¯— əy³
04% 24% 2+% 2 = 0. + axt dx²əy² dy
90. Roots of Auxiliary Equation Complex.
 If the coefficients in
the differential equation are real, the complex roots of the auxiliary equation occur in pairs of conjugates. aiẞ will also be one. mentary function will be
Hence if a + iẞ is a root,
The corresponding terms in the comple
$(y + ax + ißx) + 4( y + ax − ißx). and being any two arbitrarily chosen functions, there is no loss in putting Φ = Φι + ίψι, ψ = Φιίψι Our expression above becomes then $1(y + ax + ißx) + 41 (y + ax — ißx) + i[41(y + ax + ißx) − 41 (1 + ax — ißx) ]. For
₁ and ₁ any real functions, this is real.
PARTIAL , HIGHER ORDER THAN THE FIRST
$ 91 22%
მე 2
22%
22% 2
Ex.
243
= 0.
+2
dx dy
Əy²
The auxiliary equation is m²  2 m + 2 = 0.
.'. m = 1 ± ¿
The general solution is z = $(y + x + ix) +4(y + x − ix). It will assume a real form
ix) % = $1(y + x + ix) + 41 (y + x − + i[41(y + x + ix) — 41( y + x − ix)], for , and
any real functions.
For example, if, in particular, we choose
1 (u) to be cos u, and
41(u) to be e", we have cos (y + x + ix) = cos (x + y) cos ix  sin (x + y) sin ix = cos ( x + y ) cosh x − i sin ( x + y ) sinh x cos (y + x  ix) = cos (x + y) cos ix + sin (x + y) sin ix = cos (x + y ) cosh x + isin (x + y ) sinh x .
ex+x+ix ·ev+x ix = ev+x (eix − e−ix) = 2 ie²+* sin x. ..z = 2 cos (x + y) cosh x — :2
91. Particular Integral.
+ sin x.
 General methods for finding the par
ticular integral which must be added to the complementary function to get the general integral, in case the righthand member of the equation is different from zero, may be deduced along lines entirely
DIFFERENTIAL EQUATIONS
244
$ 91
analogous to those for linear ordinary differential equations with constant coefficients ( §§ 47 , 48 ) .*
In a large number of cases, these
can be found more simply by trial, by methods similar to that of undetermined coefficients (§ 50). The following examples will illustrate :
22% Ex. 1.
+
ax2
a2z dx dy
22%
= sin (x + 2y) — 2 sin (x +y) + x + xy.
2 dy²
The complementary function is $ (y + x) +4(y− 2 x). To get sin (x + 2y) , since all the derivatives are of the second order, we try z = a sin (x + 2 y ) . 22% 02%  2 22% = Then + = 5a sin (x + 2y) . dx2 dx dy ay² If a = }, this becomes sin (x + 2y) . lar integral is 180sin (x + 2y) .
Hence the required particu
Since sin (x + y) is part of the complementary function , there is no use in trying z = b sin (x +y) .
Trying z = bx sin (x + y) we get
3b cos (x + y) .
Hence we must try z = bx cos (x + y) . Doing this, we have as a result of substituting in the equation — 3 b sin (x +y). ― This will be 2 sin (x + y) if b = 3. Hence x cos (x + y) is the required particular integral. [ It is obvious that we might have also used z = by cos (x +y) .] To get x, we try z = cx³. Substituting this, This equals x if 6c1 ; hence the corresponding X3 particular integral is To get xy, we try z =fx³y. Using this we 6
we get 6 cx.
get 6ƒxy + 3fx².
So we try z = fx³y +gx .
6fxy + (3f+ 12g) .
This equals xy iff
the required particular integral is solution is
xy 
Using this, we get , 8= 24. 24· Hence x.
And the general
z = 4(y + x) +4 (y − 2 x) + } sin (x + 2y) + 3x cos (x + y) + ½ x³ +
zx³y — 24x² * Thus, for example, see Forsyth, § 250, Johnson, § 320 and foll., Murray, § 130."
§ 91
PARTIAL, HIGHER ORDER THAN THE FIRST
22% Ex. 2.
22%
22%
ex+2y + exty.
+2
3 dx²
245
дх ду
ǝy2
The complementary function is $ (y + x) + (y + 2 x). To get ex+2y we try zaex+2y. Using this, we get 3 aex+2y. Hence
+2 is the particular integral desired .
Since et is a part of the complementary function , let us try z = bxex+y. Using this, we get  bet . Hence — xe* + is the particular integral desired.
[ Of course, we might have used z = bye + instead . ]
And the general solution is 2= 4(y + x) +4(y + 2 x) + } ex +2x − xe²+y.
Ex. 3.
23% Əx²dy
23%
23% 2
= I • dy³ x²
+ dx dy²
The auxiliary equation is m²  2 m + 1 = 0.
.. m = 1 , I.
[We have here an example of the case cited in the footnote of § 88. ] The complementary function is p (x) + ¥(y + x) + xx ( y +x) . 23% Since there is no term in in order to get , we have to take dxs' x2 I y times a function of x which on being differentiated twice gives that is, we shall try z = ay log x. I we get x2
Doing this, we find that if a =
Hence the general solution is z = 4(x) +4(y + x) + xx( y + x) — y log x.
22%
02%
dx²
dx dy
22%
Ex. 4. 2
Ex. 5.
3
22%  22% dx² Әх dy dx ду
= sin (x +y) + 3x²y. dy
22% = xy.
2
მy2
 I
DIFFERENTIAL EQUATIONS
246
$ 92
92. NonHomogeneous Linear Equations with Constant Coefficients.  If F (D,
) of § 87 is not homogeneous in D and
, but the co
efficients are constants, it is only under certain circumstances that we can obtain solutions involving arbitrary functions, although we can always find solutions with an indefinite number of arbitrary constants. Since D'earby = arear+by, Dear+by = bear+by," the result of substituting z
ceax +by in the lefthand member of
(1)
is cear+by F(a, b) . (2)
F (D, D)z = 0
If a and b satisfy the relation
F (a, b) = 0,
which we shall call the auxiliary equation, z = ceax+by will be a solution of ( 1 ) where
is any constant.
Corresponding to any value of
b there will be a definite number of values of a satisfying ( 2 ) . Hence we can find as many particular solutions as we please by giving various values to b. Now, the sum of any number of integrals of ( 1 ) is also an integral . Hence
(3)
% = Σceax+by
is a solution where the c's and b's are arbitrary constants, indefinite in number, and each a is so chosen that with the corresponding b it satisfies (2 ) . If corresponding to any value of b we have the k values of a satisfying ( 2 ) in the form fi (b) , ƒ½(b) , … , ƒ (b) , we can write (3) in the form (4) z = Σcef1(b)x+by + Σce³½(b)x+by + ... + Σce√x(b) +by
where the c's and the b's are perfectly arbitrary. In general, F(D, ♪) has no rational factors .
If there is a
linear factor D  λD  µ, the equation ( 2) will contain the factor
$ 92
PARTIAL, HIGHER ORDER THAN THE FIRST
α  λό  µ, whence a = λό +μ.
247
Hence one of the f's, say fi(b),
becomes λ + µ, and the corresponding set of terms in (4) may be written z = Σceb (λx+y) +μx = eмxΣceb(λx+y) . (5) Since the c's and the b's are arbitrary, Σce³( + ) is an arbitrary function of Ax + y, say
(Ax + y) .
So that (5) may be written
z = eμ² (λx + y).
(6)
Hence we see that corresponding to every distinct linear factor of (2 ) we have a solution of the form (6) .* If there is a linear factor of F(D, D) which is free of D, the corresponding factor in (2 ) will be free of a ; let it be b  μ. corresponding set of terms in (4) may be written
The
% = Σceax +μy =εμυΣέραπ .
(5 ')
Since the c's and a's are arbitrary,
cez is an arbitrary function of
x, and (5 ') may be written
% = eμvp(x). емиф(х). z
(6')
If the righthand member of the differential equation is not zero, a particular integral may frequently be gotten by trial as in § 91 . Əz
22%
22% Ex. 1.
+ dx²
+ dx dy
z = sin (x + 2y) .
dy
The auxiliary equation ( 2) is
a² + ab + b  I = 0, or
(a + 1 ) (a + b − 1) = 0 .
* In the case of the homogeneous equations (§ 88) all the factors are linear. Besides, in that case, μ = o , and the result there obtained coincides with what we have found here.
DIFFERENTIAL EQUATIONS
248
§ 92
Using a + 1 = o, we have λ = 0, µ = 1 , and from (6) we see (y) is a solution. that ze Using a + b  10, we have
1 , μ = I. λ == =
Hence % =
The complementary function is, therefore,
(y  x) is a solution .
e¯* (y) + e*4(y  x). For a particular integral try za sin (x + 2y) + ẞ cos (x + 2y).
Substituting this in the lefthand member, we get
(4 α − 2 ß) sin (x + 2 y) + ( −4 B + 2 α) cos (x + 2 y). I This will equal sin (x + 2 y) if α = — B= 5 lar integral is
5
I ΙΟ
Hence a particu
sin (x + 2y) — —cos (x + 2 y) , ΙΟ
and the general solution is z = e¯*þ (y ) + e*4 (y − x) — — [ 2 sin (x + 2y) + cos (x + 2y) ]. ΙΟ
Ex. 2.
22% dx²
22% az + % = e *. +2 ay2 дх
The auxiliary equation (2) is a² — b² + 2a + 1 = 0 or
(a + b + 1 ) (a − b + 1 ) = 0.
Hence the complementary function is e¯* [$(y − x) + ¥(y + x)].
PARTIAL, HIGHER ORDER THAN THE FIRST
§ 93
249
Since e * is part of the complementary function, we would naturally try z = axe*.
But this is also part of the complementary function, y x  y + x• as may be seen by putting =2 2 We must then try z = ax²e *. Substituting this in the lefthand
Hence α = 1, and the particular integral 2 The general solution is
member, we get 2 αe¯*. desired is
a² * . 2
z { $ (( y − a) c [$(ys = e¯ x) + 4 (y + x) + =223:x²
Əz 22% Əz 22% +2 2 +2 = 2x+3y + sin (2x +y). дх ду Əy²
22% Ex. 3.
Ex. 4.
მე 2
dx dy
22% მე 2
02%
Əz
+ дх ду
z = cos (x + 2y) + e”.
ду
93. Equations Reducible to Linear Equations with Constant Coefficients. If the coefficient of D' in F (D, D) of § 87 is a constant times xy , the equation can be reduced to one with constant coefficients by the transformation log x = X, log y = Y. with Cauchy's equation, § 51.) Thus
D% =
D²% =
z Əz =I d ax x dx'
22%  I 22% მე 2 x²X²
J2%
D&z =
..yDz = уду
= I 02% ay 12 ay
22%
az
ax2
ax
.. x²D²% =
=
ду
Əz ax
I dz
x² OX'
I dz
Əz Dz
D2% =
..xDz:=
(Compare
Əz ay
I dz
... y²²% = y²ay'
22% 22% I = дх дух дХӘ ’ У у
.. xyDJz =
22% ay2
ay'
22% axay
DIFFERENTIAL EQUATIONS
250
Ex. 1 .
x2
22% 72% + 2xy dx2 дх ду
§ 93
02%
= x² + y².
Making the substitution log x = X, log y = Y, the equation becomes az 22% 22% 22% dz = e2x + e2r, which has constant co+ +2 2X2 axay ay² əx ay
efficients.
The auxiliary equation , ( 2) , § 92 , is
a² + 2 ab + b² —abo , or (a + b) (a + b − 1 ) = 0.
Hence the complementary function is
( Y− X) + e³ ¢ ( Y− X) .
For the particular integral try z = ac² + Be² .
that α = В =
•
Doing this, we find
Hence the general solution is
2
I ≈ = $( Y −X) + e³y ( Y − X ) + 1 (e² + e²²) .
Passing back now to x and y, and remembering that Y X= X = log , x the general solution takes the form
z = 4( x2 ) + x 4 (x2 ) +
(x² + 1º).
[This equation, being linear in r, s, t, comes under the head of the case treated in § 85. The student should solve this example by Monge's method, as an exercise . ]
az дх
az ду
Ex. 2.
x2
22% მჯ 2
Ex. 3.
x2
Əz 22% 22% 02% = x³yt. 4 xy + 6y + 4y² მ2 dx dy Əy² ду
22% + x.
y
$ 94
PARTIAL, HIGHER ORDER THAN THE FIRST
251
Other equations may be reducible to linear equations with constant coefficients.
(But the transformation is not always so obvious as in
the case cited above. ) Thus let the student apply the transformation I X = 1x² , Y ==y² to the following example . 2 2
Ex. 4.
I 22% x² dx²
I dz = I 02% ― I Əz y² dy² y³ əy
x³ dx
94. Summary.  The number of classes of partial differential equations of higher order than the first which can be integrated by elementary means is very small . In this chapter we have dealt almost entirely with differential equations either linear in the dependent variable and all of its derivatives, or linear in the highest derivatives only, these being of order two.
This latter class frequently
yields to Monge's method, § 85. If the equation is linear in the dependent variable and all of its derivatives, and has constant coefficients, the general method of § 92 applies. If the linear equation with constant coefficients is " homogeneous," that is, if the dependent variable is absent, and all the derivatives that appear are of the same order, the method of § 88 applies. If the equation is linear, but the coefficients are not constants, a transformation can sometimes be found to reduce the equation to one with constant coefficients ( § 93) . At times the special method of § 86 can be applied directly to an equation.
Ex. 1. ys = x +y. Ex. 2. r  s — 6 t = xy. Ex. 3. zr +p² = 3 xy². Ex. 4. xr− (x + y) s + yt = x +3(p − q)• x y
DIFFERENTIAL EQUATIONS
252
Ex.
5. xr  p = xy.
Ex.
6. r  t 3 p + 3 9 = ex + 2y.
Ex.
7. x²r  y²t = (x + 1)y.
Ex.
8. x²r + 2 xys + y²t + xp +yq − z = 0.
Ex.
9. xr+p = 9x²y².
x Ex. 10. st = "
894
NOTE I
CONDITION FOR RELATION BETWEEN FUNCTIONS
253
NOTE I Condition that a Relation exist between Two Functions of Two Variables. If u and are two functions of x and y, the necessary and sufficient condition that a relation exist between them is that their functional determinant (also called their Jacobian) vanishes, that is, ότι ax ə (u, v) = Jx,y (u, v) = ( U, V) x, y = θα J (x, y') ay
θα ax = 0. θυ ay
1° To prove the necessity of the condition . If u = (v) , differentiating, we get Ju_do av дх dv ax Judo av ay dv dy
аф These two equations in the single quantity dv can hold simultaneously only if Iau av ax ax = 0, Ju av  dy  dy 
which proves the necessity of the condition.
2° Now to prove the sufficiency of the condition. Suppose u and v to be given by u = fi(x, y), v = ƒ2(x, y) . From these we can eliminate y, resulting in a relation, which may be supposed solved for u, thus u = p(x, v).
254
DIFFERENTIAL EQUATIONS
NOTE II
33
Differentiating, and remembering that x and y are independent variables, we have Ju ЭФ , ОФ Эг + дх дх av ax' Ju = 24 av av ay Əy Iau av Эх дх If = 0, these equations can hold simultaneously only provided Ju dv ду ду is free of x. Hence , when the Jacobian Эф = 0. But this means that Эх vanishes u = $(v), which proves the sufficiency of the condition. Remark.  This theorem can be extended to ŉ functions of ŉ independent variables. NOTE II General Summary. The following is an index to the various methods, given in this book, for solving differential equations : In the case of a single ordinary differential equation, if it is of first order and first degree, see § 19; if it is of first order and higher degree than the first, see § 28 for the general solution, and § 34 for the singular solution ; if it is of higher order than the first and linear with constant coefficients, see § 52 (note what is said there of a very general class of linear equations which can be transformed to linear equations with constant coefficients) ; if it is of the second order and linear, see §§ 55, 62, and 74 ; if it is of higher order than the first and does not come under any of the above heads, see § 62. If there is a system of ordinary differential equations, see § 69. As a final resort, whether there is a single ordinary differential equation or a system of them, the general methods of Chapter XI may be tried. If there is a single total differential equation in more than two variables, see § 41. If there is a single partial differential equation of the first order, see § 84. If there is a single partial differential equation of higher order than the first, see § 94. The above may be put in the following tabular form :
first degre ,9 e 1§
solution 28 ,§ general
first order
higher degree
:
ordinary
singular solution 3,§ 4
linear with constant coefficients ,§ 2 5
System Ordinary of Differential Equations 6 ,§ 9
linear and of secon order §§42 d 55 7,6
total more than two variables 4,§in1
Single Differential Equation
higher than first order 8 4 higher 9,§the
general The methods Chapter of XI single apply ordinary to differential equations systems and these .of
s 6other ,§equation 2
partial
NOTE II GENERAL SUMMARY 255
ANSWERS
Section 12
Section 3
2 I dy 8. y = x²dx 金) ++VF( dy 4. 2 = I dx T dx2) '  [· + (2)  (23 5. x d2y_dy = 0. dx2 dx d 6. y = 2x y . dx d = 7. x = 2y y + x. dx dx
Section 8 x2 4. xy + 2 = c. 5. 3x² 2xy + x + y² − 3y = c.
Section 9 2. y + 2xy 2 x2 := cx2y. 3. y(1 + x2) = c ( x − y²). 4. secx + tany = c.
2. x2y + x³y2 := c. x I 3. log  = C. y xy Section 13
3. 2y = (x + 1 ) ª + c (x + 1 ) ². 4. y(1 + x2)2 = x² 1 + log x² + c. 5. y = x² (1 + ce *) .
Section 14 2. 3. 4. 5.
(12 — 1 ) ex² = c. (cos y 1) ecos x = c. xyt (ex + c):= 1 . Vy + 1 = x + I + c√x + Section 15
2. y2 = cx. 3. x³ + y = cxy². Section 16
Section 10 x 3. log x —— = C. y 4. 12 = cx² (x² + y²). 5. x² + y² = cx²y². 6. logx  sin = c. x Section 11
=. 3. log √x² + tan12c x 4. x2  y2 = cx.
5. logx + 12 = c. x 6. tan12 = x + c. x Section 17
1. 3x² + 8 x³y + 6 x²y² = c. + 2 + log [ 4 (x + 1 )2 2. (x² + y²) ev = c. (2x+2) + + (y + 2)²] = c. 3. xy + y² + 2x = c. 12 3. 5x  10y + log ( 10x + 5y  2) = c. 257
2. tan1
ANSWERS
258
x2 x ² + = c. 2 y x 6. x² + y² = c (my − x) . 20. =:cex+y. y Section 18 21. log r³ + logy +22 = c. 2 2Vbeza tan  ly I 22. x² +y² = c√x² +y² ™ª¯ + +½(y−x) (y x).. 4. y = Tc ke a 2Vbc a 23. x + y  4 log ( 2 x + 3y + 7) = c. x ke a +I 24. x2y2 (12  x²) = c. I if b and c have the same sign; 25. x² + y² + = c. xy x 26. √1 + x² + y² + tan−1 = c. = √; xa tan (*  √=bc a xa ), x y if b and c have opposite signs. 27. x + yey = c.
4. y = cx.
19.
Section 19
1. VT x² + VI  y2 = C. VI − x² = c. 2. XVI  y² + y √1
28. 1 = log x + 1 + cx. y 29. x2y22xy log cy = I.
30. y = ceVi
3.y + x² + 3 = ce³ . 4. Y = cely. yx  I 5. x³ ³ (3 + c) = 1. 6. xy = cex+y. x 7. logy + = c. y 8. c2x2  2 cy I = 0. 9. y = (x + a)³ + c (x + a)³. 10. sin12 = log x + c. χ
11. x sin
= c. x 12. (x + y − 1 ) ³ = c (x − y + 3). I x 13. y = + ce I  x2 x2 — a. 14. = c√1 − x² y 15. x³y² — x² = cy. 16. y tan1 x  I + cetan ¹x 7 x = c. 17. x¹ył — xîył = sin x  I + cesinx. 18. y
Section 20 2
dy +(x²−y²—\) dx —xy = 0. 2 dy 4. y dy + 2x dx y = 0. (2) dy dy 2/2)² 5. y = x² dx + r√x + ( dx 3. xy dx
dy x  y) )2 6.x(2 dx) + (スヌー刃 dx + y =a 2 7. 2x dy\2 (1/4)² dx = (3 x − 1 )². 2 = 8x³. 8. = dx (2 xdy)²= 9. x2 (d²y 2  2x dy 2/2) ²] 2 x 3yd²y[ dx dx2 1 + ( dx dx2 dy = [¹ + dx 2 d2y dy 0. 10. y dx  x ( dx)  x dx² = o @ \ =0. 11. (a) 3 ( 142) dx2 ² = dy dx day , (6) ddx3 dy =r2 I 2/2) 12 [++( dx 33)  ( dx2
ANSWERS Section 21
6. The parabolas y² = 2 λx + c. 1 7. y = ce x. 8. The circles x² + y² = cx. 9. The cardioids p = c( 1 − cos 0) . ce 10. The spirals p² := ce±0. Section 22
6. The circles x² + y² := c (my  x), (see Ex. 6, § 17) , through the origin with their centers on the line y =  mx, where m = tan α. 7. The logarithmic spirals pemo = c (compare Ex. 5) . 8. The ellipses 2 x² + y² = c. 9. The equilateral hyperbolas xy = c. 10. ya = cxb. 12. p = c( I + cos 0) .
13. peevc02. 14. pm cos m0 = cm. 15. tan ce²P. 26 16. p = the family of parabI + cos 0 olas, confocal and coaxial with the original family ( compare Ex. 4) . 12  c2 17. p = λ cos 0 C the same family of conics (compare Ex. 3) .
259
never vanishes at a finite time, while the distance covered increases continually, but never attains the value — at a finite time. k L (b) i = E + ce 6. (a) i = ce 1, R Rt L R e R SE (d) i = ce " + = E ((t) el dt. L
Section 24 — c)². 4. y2 = 2 x (x 5. 2 x = cey  I cey 6. (1 + xycy) (x + y + 1 − ce³) (x − y — 1 — ce−x) = 0. Section 25
c + e2x 4. y = x + C e2x 5. xy = c²x + c. 6. 4(x² + y) = (2x² + 3 xy + c)2. Section 26
С
x = − c ( 2 p2 + 3 ) . ²) ( 1 + p²) (1 + p²)# 2. 12  2 cx + a2c2 = 0.  c)2. 4. y = c (x — 5. The family of circles, (x − c)² + y² = c. 1. y =
Section 23
I 2. v = gt sin a ; x = x0 = gt2 sin α. 2 cert +ert. 3. x x0 = g log r2 c+ I Vo – 5. x − x0 = k (ett− 1) . If k > 0, the velocity and the distance covered increase indefinitely ; if ko, the velocity diminishes continually, but
Section 27 4. el = c (ex + 1 ) + c³.
5. y² = cx² + 1 .. C
6. x² + y² = c ( x + y) − 2. 7. y² = cx + = c³. 9. y = cx² + c²x.
ANSWERS
260 Section 28 1. 2. 3. 4. 5. 6.
(x − c)² + y² = a². cy = c²(x — b) + a. x + cxy + c² = 0 . x² + c(x − 3 y) + c² = 0. (x y + 1 )2 +2 c (x + y + 1 ) + 2² = 0. a²c y2 = cx² c+ I 7. y(1 + cos x) = c.  c². 8. (y — cx)² := 1 − 9. (y  cx)² = 4 c². 10. y² = cx² + c². 11. x2 cyex = c²xe2x. f(p)dp 12. 0 = + c. =S P√p² − [ƒ(p) ]² √142 cek 13. p = ce
14. xktan 1 +
+ c, I$ + p²)+ tan , y = kp2 ;; or putting 2 I +2 k x= (0 +sin 0) + c, y = ~ ( I — cos 0), 2 2 a family of cycloids generated by a circle of radius k We see, then, 2 that a characteristic property of a cycloid generated by a circle of radius a is that s2 = 8 ay, where s is the length of arc measured from the nearest cusp . 15. The same family of cycloids as in I Ex. 14. Here k = Section 29 1. y2 = r².
1.
Section 32 x2 + 12 I, when the fixed 2² + k k points are (c, 0) , ( — c, 0) , and k is the constant product . According ask is positive or negative , we have
2. 3. 4. 5.
the ellipse or the hyperbola, having the fixed points for foci, and +k for semi conjugate axis. The circle x² + y² = k², where k is the constant distance. The equilateral hyperbola 2 xy = a². The parabola ( x − y) ² — 2 a (x + y) + a² = 0. (y  cx)² + 4 c = 0, g. s. *, xy 10, s. s. (y − cx)² = b² + a²c², g. s., x2 = I, S. S. , a2 62 (y + x − c)² = 4 xy, g. s., = 0, s. S. xy = Section 33
4. (x− c)² = y³ —jª, g. s., y = I , S. S., y = o, c. l. , y = 3, t. 1. § 24, Ex. 3, 12 — I = 0, s. S. Ex. 4, there is no s. s. , x = 0, t.l. and p.s. for c = ∞ , y = o, n.l. Ex. 5, x² + 1 = 0, s. S. § 25, Ex. 5, 4x²y + 1 = 0, S. S., x = 0, t. l. Ex. 6, there is no s. s. = c. 1. x² + yo, § 26, Ex. 2, x²  a²y2 = 0, s. S. Ex. 4, 4x³ 27 y = 0, S. S., yo, s.s. and t.l. § 27, Ex. 2, x² + 4 e²y = 0, s. s. Ex. 6, xy = 0, S. S. , yx = 0, t.l. § 28, Ex. 2, y2 — 4 a ( x — b) = −, s. s. Ex. 3, xy²  4 = 0, S. S. , x = 0, p. s. for c = 0. * In these answers the following abbreviations are used : g. s. for general solution, s. s . for singular solution , p.s. for particular solution, t . l. for taclocus, n.l. for nodal locus, c. 1. for cuspidal locus.
ANSWERS
Ex. 4, x² + 2 xy — 3y² = 0, s.s. Ex. 5, (x + 1 ) y = 0, s. S. Ex. 11, x² + y² = 0, S. S. , x = 0, t. l. 5. 4y² = 4x + I , s. s. , y = 0, t. l. Section 36 x 2. z = cev 3. x² + y² + (3 — c) ² = a². 4. x² + y² + z² = cx. Section 37 2. xy + ya + 2x = c ( x + y + z). 3. 1 + % + 2 + x = c. x y
Section 41
1. yz + zx + xy = c. 2. x² + y² = c (x + 1)². 3. ex (x + y² + 22) = c. 4. x = Z + c. y+ a 5. ex (y + z) = c. 6. (x² + y +z) e²² = c. 7. x² + xy² + x²% − t = c. _y 8. zce *. 9. (x1) ( y− 1 ) ( z − 1 ) ex+y+z = c. 10. (z − y)² (x + 2 y) = c. 11. (y + z) ex +y = ct. 12. (y + z) (t + c) + z ( x − t) = 0. Section 43 3. y = c1 + C₂e* + cзex. 4. y = c1ex + c2e− x + cze²x. Section 44
2. y = ex (c1 + c₂x) + c3e−x. 3. y = ex(c1 + c2x + c3x²) + €4е¤. 4. y = c1 + e³x (c² + (3x) .
261
Section 45 2. y = (a + x) cosx + (c +c )sinx . x 3. y = c1 + c²² ( c₂ cos ** √3 2 x + c3 sin V3). Section 47 2. y = c₁ex + c₂е¯ 2x + e2x¿ex.
e x. x² + 201820) 60 COS x  3 sin x 4. y = c₁ex + c²²x + ΙΟ 5. y = (c1 + c2x) ex  ex log (1  x). 3. 9 = (ax +
x+
Section 48
1. y = c1ex + c2e2x . xex. 2. y = ciex + c2e− x + c³é³x x2 + 6x + 20 +9 27 3. y = c1 cos x + c₂ sin x + x sin x + cos x log cos x. 2x x 4. y = (ci + c2x) ex + cze²x  2 5. 4 Section 49 2. y = c1 cos x + c₂ sin x cos x log (sec x + tan x) . Section 50 3. y = c1 cos x + c₂ sin x + ex + x8. 7x. I I 4. y = (c1 + c2x) e− x ++ − e²x 2 sin x. 3 x cos 21+2 √3 c₁e² + e b² ( C2 5. y = Ge² 4₂ COS x x². + sin 2 √3)xt I 6. y = c1c2e−x + c3 e³x 3 I I 14 x+ X. ++ in + COS.. IO 9 5
262
ANSWERS
7. y = ( 1 + cqx) ex + ( cs + c4x) e=x I + = x²ex +4.
9. y = c1 sin 2 x + c₂ cos 2x Ixsin 2x + 8 x 10. ya cos (x + b) + 8. y = c1 + c₂e²x + ~2 (e²x − 1 ) .  1. sin x log (tan x + sec x) — 9. y = ( 1 + cgx) cos x 11. y = (c1c2x + c3x²) ex − x −3 I COS X. x2 x6ex + (c3 + c4x) sin x 120 12. y = (c1c2x + c3x²) ex + c4e2x Section 51 I + esx. 2. y = ( 1 cos log x + c2 sin log x) x 40 C3 2 log x + x + 5x + x 1 cos x + c₂ sin x 13. y x2 sin x + x cos x x 41+ log x I + + log 3. y = 4 x Ix x C3 log x 4. y = c1 (x + 1 ) ² + c₂ ( x + 1 )³ 14. y = x (c1 + c₂ log x) + x + 4x 3x + 2 x x + 6 15. y = c1ee 2 c2 cos √3 ( xex Section 52 COS X  sin x 2 + 6 (x  3) +ca sin√3) 1. y = c1e2x + €₂e³× + IO cos 2x + 8 sin 2 x I + xe2x. 2 130 I 2. y = c1ec2e− x + c3 sin x + c4 cos x ex x sin 16. y = ( 1+ (2x) — I  ex cos x. 5 + ex 3. y = ( 1 + c2x) e− x + 2 x³ − 12 x2 ( +1) 1°. k > n. 0 = ekt(AeMt + Beμt), 18. (2 x − 1 ) e³x + 36x  48where μ2k2² — n². • diminishes 32 continually but vanishes only for x4ex 4. y = (c1 + cqx + c3 x²) 6ex + t = ∞ , theoretically. Practically, 24 x3 x the swing ofthe pendulum is soon 5. y = c1 + c2e2x + cze−2x 12 damped sufficiently, so that the assumed law ceases to hold, and ℗ be 3x²x. comes zero in a finite time. But 6. y = (c1 + c2x) ex + ( c8 + c4x) e −x the solution tells us that the pen+ =I cos x. dulum comes to rest without vibrat4 ing. 7. y = ( 1 + c₂ log x) sin log x + 2°. k < n. 0 = Aekt cos (ut + B), (c8 + c4 log x) cos log x + (log x)² where u² = n² – k². The pendu+ 2 log x 3. lum vibrates with constant period x3 5x2 8. y = c1 + ( c2 + c3x) e−x + 2π 2 3 which is longer than + 8x. Vn2  k2
ANSWERS
263
in the case of motion in a vacuum for case (a) . If k is small, the (Ex. 17) . The amplitude is Aekt, amplitude is large in case (b) , while which diminishes continually. in case (a) the amplitude increases with the time. This explains reso3°. k = n. 0 = (A + Bt) e¯kt. If A and B have the same sign, nance in Acoustics, the effect of diminishes continually, as in case the measured step of soldiers on a 1º. If A and B have opposite bridge, and the like. passes through zero, 20. x= xocos kt+ sin kt. The period signs, k changes sign, attains a maximum is independent of xo and vo in absolute value and then diminishes continually in absolute value, 21 . 2x= = ( xo + 20) k ett + ( x0  20) k e kt. as before. 19. (a) 1º. m ‡ n. 0 = A cos ( nt+ B)  22. r = c1ewt + cze  wt + 2 62 sin wt, where C COS mt. + n2 m2 c1 + c₂ = ro, w(C1 – C2) + _g_ = v0. 2w 2º. m = n. 0= A cos (nt + B) If c₁0, the first term soon predomiCt nates, and the motion is spiral. + 2n sin nt. If 10, the second term soon (b) The same complementary funcbecoming negligible, simple harmonic tion as in Ex. 18 * motion results. T + C [ (n² m²) cos mt + 2k sin mt] When roo and v0 = & , (n² — m²) ² + 4k² 23 sin wt ; i.e. we have simple This last term may also be written  r = w2 2 C cos (mt a) , harmonic motion from the beginning. √(n² − m²)² + 4 k² where tan α = 2k Section 53 The part n2  m² + c²x²ex + x. 2. y = c1e of the motion indicated by this term is called the forced vibration. 3. y = c1x + c2 ( 1 + x tan−¹ x). 2π Its period is It is not in 4. y = c1x + c2ex + x² + 1 . m 6. y = sec x(c1eax + c₂e  ax) . phase with the periodic force [except in case (a) 1 °] , lagging be 7. y = x²e 4 (c1 + c₂ log x). hind by the angle α. When 8. xy = c1ex + c₂e− x + xex. π n > m, o < α < when Section 54 π when nm , n < m, 2 2. y = c₁x√1 − x² + c2 ( 1 − 2 x²) . α = π. In this case the forced 3. y c₁ sin (sin x) + c₂ cos (sin x) . 2 I vibration is given by U = C sin nt 4. y = c1 cos I + c2 sin I + 2k 2x2 2 x2 x2 Ct sin nt I for case (b) , and by U = 5. y = c1e2x² + c₂ex²  x²ex². 2n 3 * Replace k by km throughout the answer.
ANSWERS
264
Section 61
Section 55
1. y = c1ex + c2 (x² + 3 x² + 6x + 6). 2. y = c1ec2e³x (4x³  42 x2 + 150x  183 ). 3. y = (c1ex + c₂e*) . x2 4. y = c1 (x²  1 ) + c2x. 5. y = ex(c1 + c₂ log x) . 6. y = x² (c1 cos x + c₂ sin x) . 7. y = c1x + c2(x² + 1 ) . 8. 2y = x(c1e2x + c2 − x) . 9. y = c1x cos (ax + c₂) . n n 10. y 1 cos + c₂ sin x · x
C1x I + (2x 3. logy = c1e + c2e− x + x² + 2. 4. y = c1 cot x + c₂ ( I  x cot x). 2. y = x log
Section 62
1. y c₁log cos (x + c2). 2. y = (sin1x) ² + c₁ sin¹ x + c². 3. a + Ly = ea(x+b). ay 4. ( 1 + x³) y = c1x² + cqx + €3. + 2x5. ( x − 1)5y = c1 ( x¹ — ·6x² 6 x² +
3 Section 57
2 dx + €2. 3. y = x + q Se 4. y = (x 2) e * + c1x + €2. 5. y = = c1x² + x√1 + q² +c2• Section 58
 a)² + y² = b. 2. (x − ecx 3. ey = 4 ac²  αecx)2 (1 − 4. ay + beax + 2 = 0.
 4x³log x ) + cx³ I 6. logy = 1 + C1x + c₂ 7. y = c₁ log x + c2. 8. x2y + xy² = c1x + C2— 12 9. y = log cos ( 1 − x) + €2. x2 + c₁√x²  1 + C2. 2 11. y = c1 + (c₂ + c3 log x) √x. 12. y = c1 sin2 x + c₂ cos x c2 sin² x log tan . 2
10. y =
13. (a) The circles (x + c1) ² + y² = c2. (b) The catenaries x+ 2 y: £1 e 1 te 91 2
Section 60 x2 2. xy = + c₁ log x + c2. 4 14. (a) The cycloids 3. (x  1 ) 2y = c1 + c2x COS X. x + c1 = c2 vers ¹Y — √zczy—y², 4. xy√x²  1 = q√x²  I C2 + c₂ log(x + √x² − 1) + cs. (b) the parabolas 5. x³y2 = c1x² + c2x + c8. (x + c1)² = 2 c2y — cq².  c₁ sin¹x √ — x² 7. xy = c1x — 15. The central conics  x². + c₂√1 − 6132 — k 12 (x + 2)² = 1 ; hyperbolas 8. y = c1 cos x² + c₂ sin x² + c3x². ellipses when k < 0. whenko, 10. tan y = c1 cot x + c2.
ANSWERS
16. y = I 2
c₁ (a  x) II n
n
1+ I (a  x) + c2, if n + 1, C1 1+1 n y= (a (a − − x) + €2, — x)2´ x)² — C¹log(a 2 4 C1 if n = I. de 17. v= l = √2 gl (cos 0 cos α). dt 18. (a) v² = 2 k² 24 ( 1), I a ax x2 (b) t= √ 2 a vers12a + ), +
(c)2/2
I I (d) v² = 2g R² R ´R+h) R +h ; if h∞ , v = √2g R, which is 7 miles per second, approximately. Section 64 1. x = ‹₁e³² + c2e−³¹ — 6 t,
y = − 2 c₁e¾t — c2e−³¢ + ² et +9 7+ 9.
2 c1  2 cat)et I  9 3 I y = (c1 + c₂t) et + c3e 3 = 1½t. 2 3. x = ( 1 + cat) et + (cs + cat) e , 2y = (c₂  c1  cat)et − (C3 + €4 + C4t) et. 2. x= (6 c2 I
Section 65
6. x³  y³ = (1 ,
x²  12 = c2.
7. x² +12 = 41,
t= C2x + y. x  czy
265
8. x²  1² = c1, = Cg. 2 (x + y) 9. y = cit, x² + 12♪ + 12 = c2t. 10. x  y = c1 (x − t) , (x +y + t) (y  t)² = cq. 11. xy  t = c1, x² + y² — c₂t² = 0. y x 12. x³y³t = c1, 202 + = C2 Section 66
x² + z² = c1, y² + z² = €2. Section 69 1. The path is a parabola lying in the vertical plane determined by the direction ofthe initial velocity. Taking the initial position for origin, the horizontal line of the plane through it for the x axis, and the vertical line through it for the y axis, the equations of the path are I x = vot cos α, y  votsin a  1gtz. 2 Or, eliminating t, we have gx2 y = xtan α2 v02 cos² α 2. x = Vo cos α (1 − e−kt), k y = kv sink2α + g ( 1 − e−kt) Σt. k 3. xa cos kt + Vo cos a sin kt, k v sin a sin kt. y= k Or, eliminating t, we have a²k²y² =a² sin² α, (xsinα y cosα) ²+ 02 V an ellipse with its center at the origin. Since x and y are periodic 2π functions of period we see that k" the motion is periodic, the period being independent of the dimen sions of the ellipse.
266
ANSWERS
4. 2kx =·(ka + vo cos α) ekt + (ka − v cos α) e kt, 2 ky = v sin α (ekt − e−kt ) . Or, eliminating t, we have a²k2y2 (x sin α y cos α)² – v02 = a² sin² α, a hyperbola with its center at the origin. 5. pa sin (kt + b), q = acos (kt + b), r = c, where a Vpo² + 90², A b = tan110 , c = k = ro A 90 Angular velocity = w = √po² + 90² + r², a constant. Direction cosines of the instanta
or y = A (x²  5x + 10−10 x−1 + 5 x−2 I x8) − x¯³) + +B B(( xx−¹ — 2 — x¯² + ΙΟ —— x˜³ ). 25 4. y = A ( x + 2.5 + 2. 4.5.9 x7 + 24659.13 + ...) x2 x4 ²( 1 + + + +B Bxx¹³ + 2.3 2.4.3.7 x6 + 2.4.6.3.7.11 + ...) x2 x4 + + 1.3 1.3.3.7 x6 + 1.3.5.3.7 . II 5. y = A (6  4x + x²) + B(x24x1). Section 75
neous axis of rotation are 1, 2,
. W 6. The paths are the curves of intersection of the hyperbolic cylinders x2 z2 12 22 = (1,  == c2. 62 c2 a2 c2 The time is given by ო 3
t to =
dz ab S% √(@² + & c ) (3² + c² (3 ) +
For the curve through the origin C1 = C2 = 0. Hence the path is one of the lines x= a C The time in covering a part of it is
t  to = ab
=
土 /  1x) · +1 ca (1 − 1 ) = +bc (xo
Section 74 I I AI 3. y = 4 4 ( ΙΟ ) + B(x * 2x+ 1
5. F ( 1 , 1 , 1, x ), n N I I 2 2· F ( − 1), 1— , xx² ) , 2 — " —2 1 2 n 2 nxF 3, x²), x F( − "— 2 1 , − "2 + 1, 2 I Limit α, β= F (a, B, 8 4 αβ).
Limit a, B=∞ 2 F ( α, B
)
Section 76 x− y + z (p − q) = 0. xqyp = 0. z = px + qy + √p² + q². z =pq. ( 1 + p²)² (p² + q² + 1 ) = r², (I + g²)² (p² + q² + 1 ) = 1², p²q²(p² + q² + 1 ) = s². 6. z =·xp + yq· xys, r = 0, t = 0. 7. r = 0, s = 0, t = 0.
1. 2. 3. 4. 5.
Section 77 − 5 x² + 10x 1— 10 + 5 x − x² ),  2. xzp + yzq = xy.
· ANSWERS
267
3. yp  xqy2 — x². I 16. z = x + exy (y) . 4. r t= 0. y + 5. xr – p  9) . 17. z = y sin−1 + − (x + y )s + yt = x = 3 ( b. + $(y). y Section 79 Section 84 I I 3. (xy + z, x² + 12) = 0. 1. Φ = 0. x y y 2 4. ø (x² + y² + z², x + y + z) = 0. 2. z = ax + (a² + 1 ) y + b. Section 82 Z 3. xyz  3u = $ 9 x x 2. y=(b  x)(a + yz) .* 3. 62 (2x  a) ³ + 6 a²y + b. 4. z = ax + by + (a + b)². 1 Section 83 5. z = cxa ya. 3. e² = cxayla. Since the equation is 6. x2. ·z² = 0 (x³ — y³) . I linear its general solution is 7. z = axe¥ + — a²e² + b. 2 ez = xf\ 8. VI + az = 2Vx + ay + b. ( ) 4. % = cxay√1  a². 9. z = a √x+ y + √1 − a² √x − y + b. 5. z = ax + a log y + c. Since the 10. z² a²x² = (ay + b)². equation is linear, its general soluI tion is z = f(x + logy) . 11. z = 2 a log (x² + 12) cxeay. 7. az ― I = + VI  a² tan12 + c . 8. 4 az = (x + ay + b)². x 9. z = ax + a²y² + b. = 12. x² + y² + 2² 20 10. log z = (x + a ) ² + ( y + a )² + b. (  a² = 0. 11. z = ax + by + √a² + b² + 1 . Sin 13. ( a − 1 ) z ( x + ay + b) — gular solution is x² + y² + z² = I. 14. =: % axy + a² (x + y) + b. Z 12.  == log (x + y − 1 ) + y + b. Since 15. xz + x √x² + y² + z² = $  a the equation is linear, its general  16. 2 az = (x + ay)² + b. solution is 17. (xu)3(x + y + 2 + u) log (x +y  1) + y =f(z) . x ν ν a)eay. + 13. z = c (x + y : " 2  u 2 u y 15. 22 = x²+(y) + 18. The family of planes (1)
*Attention should be called to the fact that no unique answer can be given for the complete solution. Other forms just as good as the ones here given may be found by selecting various forms for the auxiliary function .
z = ax + √k² − 1 − a²y + b ; for b, a fixed constant, the corresponding planes envelop the circular cone (k² − 1 ) (x² + y²) = (2  b)2, whose vertex is at the point (o, o, b).
ANSWERS
268
This equation is a particular solution for all values of b. 19. The surfaces of revolution x² + y² = $ (2). 20. The family of planes z = ax+ by+ √ (k² + c² ) a² + k²b² + k², where the fixed points are (c, 0, 0) and ( c, 0, 0) , and the constant product is 2. These envelop the ellipsoid of revolution k²x² + (k² + c²) (y² + z²) = k² (k² + c² ) .
5. z = p (y + 2 x) + 4 ( y − x) + 12 — (2 x³ + µ³).
Section 92 3. zyx) + e− 2x4 (2x + y) 1 ¿2x + 3y — — cos (y + 2x) . ΙΟ 4. z2 = exp(y) + e− x¥ (x + y) +1 2 sin (x + 2y) — xe".
Section 85 4. x = $(z) + ¥ (x + y + z). 5. x = p (z) + 4 (1') . 6. y = xp(ax + by + cz) +4 (ax + by+ cz) . Section 86 xys 3. z = ·+ y²p(x) +4 (x). 3 4. % = x212 + $(x) + ¥ (y). 4 5. z = x2y  xy + (y) + ex¥(y). 2
Section 88
Section 93
4 (xy) + 2. + (~x'.) · +* = 4(xy) 2. =2 =
3. z = p (x²y) + x¥ (x²y) + x³yª. 30 4. z = p(x² + y²) + 4 (x² − 1²).
Section 94 x2 1. z = logy + xy + 4 (x) + ¥ (y) . 2 2. z = (y + 3 x) + 4 ( y − 2 x)
1. z = p(y + x) + 4 ( y + 2 x) . 2. z=41( y) +2( y + 5x) + $3( y + 2x). Section 89
1. % = (y + x) + x¥ (y + x) . 2. z = 41 (y + x) + 42 ( y − x) + x3 (1 x). 3. z = 1 (x + y) + xp2 ( x + y) + $3 (x − y) + x$4 (x − y) .
3. 4.
5. 6.
+ = x³y + 1x4. 24 z² = x³y² + xp(y) + ¥ (y) . 2% = p(xy) +4 (x + y) . I z =  x²y log x + x²p( y ) + 4 (y). 2 2 (x + 1) + e³x (y  x) ye +2y.
7. 2=4(xy) +xy (2) + (x − 1 ) yloga.
Section 91 4. 2 = (yx) +
(2y + 3x) I + ² sin (x + y) + ¦ x¹y + 80x5.
8. == ¹ ( 2 ) + x+ (2)9. = x²² + (y) log x + (y). 10. z = ( x) + ( x + y ) + (x + y) logy.
INDEX The numbers refer to pages. The following abbreviations are used : c. c. constant coefficients. d. e. differential equation. 1. linear. o. ordinary. t. total. p. partial.
Additive constant , 2. Darboux, 75, 202. D'Alembert, 233. Degree of d. e. , 2. Applications, 3148 , 55 , 60, 67, 68, 74, 117 Derivation of o. d . e., 3. 122, 146148, 162, 163 , 214, 229. of p. d. e., 196201. Differential equation, I. Arbitrary constant, 2. of a family of curves, 31. Auxiliary equation, 92, 240, 246. ofsimple harmonic motion, 118. Discriminant, 65. Bernoulli, 20. Bernoulli's equation, 20. relation, 65. Cauchy, 91, 92, 113, 164, 202. Envelope, 62, 63. Essential arbitrary constants, 3. Cauchy's linear equation, 113, 178 . Euler, 25, 91, 192. Cayley, 75. Characteristic equation , 92.. factor or multiplier, 25. Exact differential , 8. Charpit, 218. d. e., 8, 11, 137. method of Lagrange and, 215. Existence theorem for o. d. e., 164. Clairaut, 25, 56. Clairaut's equation , 56, 59. for p. d. e., 203. extended, 225. First integral, 52, 230. Commutative operations, 97. Functional determinant, 253. Complementary function, 91 , 239. Complete 1. d. e. , 90. Gauss, 192 . Complete solution of o. d. e., 5. Gauss's equation , 193. ofp. d. e. of first order, 213. Condition for exactness of o. d. e. of first General integral, 6. order, 9. General plan of solution of o . d . e. of first order, 9. for exactness of 1. o. d. e., 138. for integrability of t. d. e. , 77. of higher order, 131. General solution of o . d. e. , 5, 167. for relation between functions, 253. of p. d. e., 203. for repeated roots of an algebraic equaof p. d. e. of first order, 213. tion, 65. Gener al summary, 254. Consecutive points , 70. Geometrical significance, 31 , 6163, 66, 69Curve of pursuit, 147. Cuspidal locus, 70. 71, 86, 157 , 167, 203, 204, 213. 269
270
INDEX
Homogeneous, function, 14. 1. o. d. e. , 89. 1. p. d. e. of first order, 205. 1. p. d. e. with c. c. , 239245. o. d. e. of first order, 14. t. d. e. , 81 . Hypergeometric series, 194.
Integrable t. d. e., 7684, 86. form of solution , 76, 85. method of solution, 80. Integral, 6. curve, 31. surface, 86. Integrating factor, 8, 25, 141 , 209. by inspection, 23. of o. d. e. of first order, indefinite in number, 8. Integration in series, 164195. of o. d. e. of first order, 169. of o. d. e. of higher order, 177. Intermediary integral, 231.
Linear p. d. e. , of second order, 230. reducible to equations with c. c., 249. Linearly independent functions, 90. Liouville, 142.
Monge, 230. Monge's equations, 231. method , 230. Nodal locus, 69. Nonhomogeneous 1. p. d. e. with c. c., 246. Nonintegrable t. d. e. , 85, 86. Nonlinear p. d . e. of first order, 211226. Order of d . e., 2. Ordinary d. e., I. of first order and first degree , 730. of first order and higher degree, 4975. of higher order, 89–148. reducible to 1. d. e. of first order, 20. system of, 149–163. Orthogonal trajectories, 38, 158.
Jacobian, 253. Kowalewski, 202. Lagrange, 75, 103 , 205, 213, 215, 218. method of, 205. Lagrange and Charpit, method of, 215. Legendre, 114. Legendre's linear equation , 114. Leibnitz, 25. Linear o. d. e., 89. Cauchy's, 113, 178. complete, 90. general, 89. homogeneous , 89. Legendre's, 114. of first order, 18. of second order, 123130. reducible to equations with c. c. , 113, 114, 126, 127. simultaneous, with c. c., 150. with c. c., 91122. Linear p. d . e., general , 238. " homogeneous ," with c. c., 239245. nonhomogeneous, with c. c., 246. offirst order, 200, 205.
Partial d. e., I. offirst order, 205229. of higher order, 230252. Particular integral of o. d. e., 6. of 1. o. d. e. in general, 103105, 125. of 1. o. d. e. with c. c., 97113. of 1. p. d. e. with c. c., 243, 247. Particular solution of o . d . e., 5. of p. d. e. of first order, 213. Picard, 75, 164. Point of general position, 32. Primitive of o. d . e. , 4. of p. d. e., 196, 199. Quadrature, 6. Reduction of d. e. to equivalent system, 160. Regular function, 203. Riccati, 173. Riccati's equation, 28, 173. analogy to 1. d. e. , 176. Roots of auxiliary equation repeated, 93, 240. complex, 94, 242.
INDEX
271
Separation of variables, 13 , 25. System of o. d. e. of first order, 153. of t. d. e., 159. Series, integration in, 164195. Singlevalued function, 165. Taclocus, 71. Singular point of an equation, 203. Singular solution of o. d. e. , 63, 6675, Tangent, 70. 168. Total d. e., 7688. method of solution, 8085. of p. d. e. of first order, 213. Solution of d. e. , 2, 164, 202. more than three variables involved, 83. simultaneous , 159. Summary, general, 254. Symbol D, 89, 238. Trajectory, 38. , 114, 238. Ultimate points of intersection, 61. Symbolic operator (D− a) , 96. Undetermined coefficients, method of, 107. System of d. e., 149–163. general method of solution , 149. Variation of parameters, 103. of 1. o. d. e. with c. c., 150.
REFERENCES
D'Alembert, 233. Boole, 101, 169, 174, 230. Cauchy, 113. Cayley, 75. Chrystal , 75. Craig, 91. Darboux, 75. Fine , 75 . Forsyth , 174, 230, 244. Gauss, 195. Glaisher, 75. GoursatBourlet, 202, 203, 213.
Hermite, 91. Hill, 72. Jackson , 122. Johnson, 174, 244. Lie, 25, 125. Liebmann, 75, 168, 169. Lobatto, IOI. Marie, 233. Murray, 164, 244. Picard, 75, 164, 165, 169, 202. Schlesinger, 125, 141 , 164, 183, 185, 194. Tait and Steele , 45.
d ( y) = xdy =2 y dx χ
= yda  xdy. d(y)
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) 2/4"
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こ
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HART'S
ALGEBRAS
The widespread fame of this series of Algebras, and its adoption by states, cities, and towns for use in thousands of schools is convincing proof of the authors' wisdom in planning a course that contains a unique progressive treatment of the equation and its applications, and reserves for a later discussion the more abstract generalizations and the more difficult type forms, thus furnishing the largest amount of usable Algebra that has been brought within the limits of the first year's work.
FIRST
YEAR ALGEBRA
Contains sufficient material for a strong first year course. Eminently teachable ; emphasizes the parts of Algebra of use in business and industrial life ; has practical, concrete problems ; with ample drill upon factoring and the use of the equation as an accurate and timesaving mathematical aid.
NEW
HIGH SCHOOL ALGEBRA
Contains ample material for standard high school courses, and includes all the topics ordinarily required for entrance to college. The first 275 pages are identical with the authors' First Year Algebra, and the book provides a continuous and consistently developed course.
SECOND
COURSE
IN ALGEBRA
Designed for a second course which is separated from the first by the lapse of a year or more. Chapters I to VIII comprise a brief but complete review of the rudiments ; Chapters IX to XVIII duplicate corresponding chapters of the New High School Algebra; and Chapters XIX to XXIV contain new material designed to enable schools to meet the highest forms of college entrance requirements.
D.
C.
HEATH
&
CO . ,
PUBLISHERS
UNIVERSITY OF ILLINOISURBANA
504
3 0112 065944883
51 C6
20