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dU

dU which does make = 0); and as it does not make

dp

dy

vanish, it is a singular solution.

(10) In the same way it will be seen that

yo + (x − 1)= 0 is a particular integral of

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If the differential equation be of an order higher than the first, let , Yu...y, represent the successive differential coefficients of y with respect to X. Then

U = 0 being the equation cleared of fractions and radicals as before, the conditions that Yn-m= X should be a singular solution of the (11 m)th order are dU dU

dU 0,

dyn dyn-1 dyn-m+1 and therefore if we find a relation between x, y, and Yn-m = X, which satisfies these equations and also the given equation, it is a singular solution of the (n m)th order.

Legendre, Mém. de l'Acad., 1790, p. 218.

=(...

= 0;

2

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Y2 we may

dạy

dy d’y (11) Let

2

+ x = 0 d x dx

dy dy be the given equation. Putting

dx

d x2 write it

x Y," 2yı Yu + x = 0.

dU Here

2 (ir Yy) = 0

dY2

C

gives yz = ""; and, by means of this, eliminating y, from the original equation we find

yi? - x2 = 0

dy

dy2

Y2 =

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as a singular solution of the first order. As this does not dU

DU satisfy = 0, and as there is only one factor in 0, there is no singular solution of the final integral. (12) Let dy

(dy
y - X +
dx 2 d x2

= 0.
dx dx

dx
dU
The condition = 0 gives us

dyz

4xyı + m2

4 (1 + x)' from which we find the singular solution of the first order to be

(dy

+ XC
d.
1+ -y 1

= 0.

16
(13) Let the equation be
dạy dy (dy)

dy)
-y +x
d x2

=0. d.r

- xy y + x \dx

dx It will be found that

XY - 1 = 0 dU dU satisfies

0, = 0 and U = 0, and as it is independent

dy2 dy of y, and yit is the singular solution belonging to the final integral. (14) If the equation be of the third order

'dy
+
- 4

X +

= 0, dx3 dr

dx a singular solution of the second order is

d’y

+ x = 0.

d02
(15) Let the equation be

dy
- 4

+

= 0. \da.

dx2

3

a?

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Then yz + x =0 is not a singular solution, because though
dU

dU
it make = 0) it also causes to vanish; and we find

dy.

dyz

=0

that the real value of is infinity instead of zero, as it

dyz would be in the case of a singular solution. Having given the general integral of a differential

equation of the first order, to find the singular solutions of the equation when there are such.

Let u = 0 be the integral cleared of radicals. As it is supposed to be an integral of an equation of the first order, it must contain an arbitrary constant, which we shall call c. Then if the equation

dU

dc give a value of c in terms of u and y, the elimination of c

dU between U = 0 and 0 will give an equation in x and y,

de which is the singular solution. It is to be observed that if dU

o give a constant value for c, or a value in terms of x de and

Y, which becomes constant in consequence of the relation U = 0, the result of the elimination is not a singular solution but a particular integral. (16) Let the equation be

.22 - 2cy - C - a' = 0.

dU Then

2 (y + c) = 0,

dc whence e = so that Q2 + y - a' = 0 is the required singular solution.

x?

62 (17) Let y - ax

? + c - 0

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=

Y,

dU

Then

:b? (c a)3 – 2? (c a) = 0.

dc

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This is satisfied either by c= a or c= a +

b The former gives a particular integral. The latter gives

46% (y ax) – x* = 0, which is the singular solution. (18) Let (wo? + yi a) (yo 2cy) + (x2 - a) c = 0.

dU Then

de = 2 {c(x* – a') – y (x2 + yo – a')} =

y (x2 + yo a*) from which

w2 a" which being substituted in the equation gives

202 + a = 0 as the singular solution : but since this makes c = 0,

o, it appears that it is only a particular integral found by making the arbitrary constant equal to zero.

Let U = 0 be the integral of an equation of the second order, so that it contains two arbitrary constants Ci, Cy; then, if we represent the differentiation with respect to x and y by d, and that with respect to e, and c, by d', we can obtain the

dc, singular solution by eliminating Ci, C2, and between the

dc,

equations

U = 0, dU = 0, d'U = 0,

d'U = 0, dd'U = 0.

(19) Let the given integral be

y = cx + Cg x + ;? + cz?, so that

U = } Ca? + C2 x + + co? - y = 0.
Then dU = (C, X + cx) dx dy = 0,

d'U = (} x + 2c,) dcı + (x + 2c2) dc, = 0,

ddU = (xdc, + dc.) dx = 0. From the last we find

dc, dc,

Substituting this in the preceding equation we have

4 (C, - c, x) - r = 0, Between this equation and the first two we can eliminate C

and and we find as the singular solution of the first integral dy

dy

x
+ ( x + 1) * - y (1 + xrP) = 0.

16

و2)

dx

2

(20) Let the integral be

y = }æ? + c) x + C C2 By a similar process to that in the last example, we find as the singular solution belonging to the first integral of the differential equation xo

- 4xy = 0.

dix There is no singular solution belonging to the final integral, but the singular solution just found has itself a singular solution, which is

x + 2y = 0.

dy

+

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+ 3x

4

Singular Solutions of Partial Differential Equations.

If U = 0 be a partial differential equation of the first

dz

dx order in X, Y, and and if we put P,

-9,

the

du dy singular solution, if there be one, will be found by eliminating p and q between the three equations

dU

dU U = 0,

0,

= 0.

dp

dq

(21) Let the equation be

(z - px - qy)2 = a' (1 + pp + q°).

dU Then

= -(a

- px - Ty) x aʻp = 0,

dp

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