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dp

p(y

+ p) = 1,

dy multiplying by y and integrating, we have

(29)° = a* + y;
whence X + c = (a® + y).
dy

dy
- Y
dx

dx

2

(33) Let

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dy dp

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dy

P - y

dy which is of Clairaut's form. The general integral is therefore

p= Cy + n(1 + a'C?),
whence + C' = log {Cy+n (1 + a'C')}}.
The singular solution is

(7°°° - y)^.
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The examples in this section are taken chiefly from Euler, Calc. Integ. Vol. 1. Sect. II. and Vol. 11. Sect. I. Cap. 2 and 3.

CHAPTER V.

INTEGRATION OF

DIFFERENTIAL EQUATIONS BY SERIES.

(1) Let

and a r" y =

The method employed for integrating Differential Equations by series, is to assume an expression for the dependent variable in terms of the independent variable with indeterminate coefficients and indices, and then to determine them by the condition of the given equation.

day

+ ar y = 0.

dir? Assume y = ma (A + A, 2" +2 + A, 22n+4 + Az xr2n+S + &c.) Whence we find dy

= a (a-1) 4.2-2 + (a+n+ 2)(a +n +1) A, 7+" + &c. d x2

a Awa+n + a A, xa+20+2 + &c. Substituting these values in the equation, and equating to zero the coefficients of the powers of x, we have a (a – 1) A = 0, (a + n + 2) (a + n + 1) A, + a A = 0),

(a + 2n + 4) (a + 2n + 3) A, + a A, = 0, &c. The first of these is satisfied either by a = 0 or a = 1. Taking a = 0) and substituting it in the other equations, we find a A

a’A A (n + 1) (n + 2) 1.2 (n + 1) (2n + 3) (n + 2)? ? a A

&c. &c. 1.2.3 (n + 1) (2n + 3) (3n + 5) (n + 2) so that

ax?n+1 y = A {1

&c.} (n+1) (n + 2) 1.2 (n + 1) (2n + 3) (n + 2)?

Az

ar+2

+

we

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+

But as this contains only one arbitrary constant 4, it is not the complete solution. Let us take a = 1 and call A', A1, A,', &c. the corresponding coefficients ; then find in the same way as before

a' xx+5 y = A' {x(n+3) (n + 2) 1.2(n + 3) (2n + 5) (n +2)

- &c.}

2 which is another incomplete integral with one arbitrary constant. The sum of these two series is the complete integral of the equation.

When n = 2 both the series fail, as the denominators are then infinite: but the true integral is easily found.

d'y ay For if

= 0, d x2

+

a.

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and we assume y= A wa, we have

a (a - 1) + a = 0. This is a quadratic equation, which gives two values for If these be an, a, the integral is y = 4,241 + A, ",

2r-1 The first of the preceding series will fail when n=

(2r + 1) and the second when n =

r being any whole number: the complete integral may however be found by the following process. Assume

y = u + v log cæ, where v is the particular integral furnished by the series which does not fail. On substituting this value of y in the original equation we obtain the system of equations

dv

+ a " v = 0, d x2

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2 do

da u d x2

+

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V

+ aua" = 0; x?

a dx

Euler, Calc.

the second of which serves to determine u. Integ. Vol. 11. Chap. vii.

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Assume U = B + B, X + B x + B2 ix} + &c.

A
Then we find B ; B, is left undetermined and

a

B2

a’B ,

За А a B
13.22

1.2'
- 5a A aA
B2 =

+
12, 23, 32 1.23 1.22.3'
&c.

&c. But since we have introduced the arbitrary constant c in log cx, we may assume for B, the value zero, and then we have

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(4) From the equation

d’y dy

+ + y = 0, d x2

dx we easily obtain a particular integral. For if we differentiate the equation r times, we have d'+2 y

drtly d'y
+ (r + 1) + = 0,
dx+2

dx+1

dict and when x = 0

dr+y

1

d'y
dw*+1
Thus

any

one of the coefficients in Maclaurin's Theorem is derived from the preceding one.

Let the first coefficient, or the value of y when x = - 0, be A, then find the particular integral

002 23 +

+

&c.) 12 12. 22 12.22.39 12. 2o. 3o. 42

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we

as

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y= A (1

Let this be put equal to v: then assuming

y = u + v log ca,
and U = B + BX + B + Bzx3 + &c.,
we find by substitution in the given equation
3.02 11.03 51 r*

B
U = 2 A (20 +

+ &c.) +
2'. 33 23. 33. 43

A

Hence we have

3.x 11x03 51004 y = 2 A (v – +

+ &c.) 23

23. 33 23. 33. 43 X2

2013 + (1 +

+

&c.) log cx, 12 12. 22

32 12. 2?. 32. 42 B the constant being included in c.

A

Fourier, Traité de la Chaleur, p. 372.

12. 2?.

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