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dU

which does make = 0; and as it does not make a vanish, it is a singular solution.

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

y + (x - 1)? = 0 is a particular integral of

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- + x = 0.

d x) dx If the differential equation be of an order higher than the first, let 91, Y2...Yn 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 (n m)th order are

du - dUdU

-= 0, dyns

1 = 0 ... dyn-1

= 0;

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.

dy)? dy d'y (11) Let « (

\d x)

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gives yz = 9; and, by means of this, eliminating yz from the original equation we find

yi? – 2.2 = 0

dU

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as a singular solution of the first order. As this does not satisfy = 0, and as there is only one factor in = 0,

dyz there is no singular solution of the final integral.

(12) Let

g-orang tema - -
The condition 10 = 0 gives us

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4xY, + gula

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

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It will be found that

wy - 1 = 0 : du du satisfies = = 0, = 0 and U = 0, and as it is independent

S dyzdy, of y, and y, it is the singular solution belonging to the final integral.

(14) If the equation be of the third order

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INI

Sada

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dn2

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value old 43

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

dU it make 40 = 0 it also causes to vanish; and we find

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

= 0

de 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,

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

= 0 give a constant value for c, or a value in terms of a 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 - a2 = 0.
dU

= – 2 (y + c) = 0,

dc whence c = – y, so that a + y - a2 = 0 is the required singular solution.

Then

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Then

dU
dc

= b(c u)3 – 2? (c – a) = 0.

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This is satisfied either by c= a or c= a .
The former gives a particular integral. The latter gives

46° (y ax) – W' = 0, which is the singular solution.

(18) Let (w* + yê – a') (2cy) + (22 – a®) c = 0.

Then 40 = 2 {c(x® – a®) – y (x2 + yje – a®)} = 0; from which

y(x* + yo a)

2 - a which being substituted in the equation gives

mo? + a= 0 as the singular solution : but since this makes c = 0, 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 cı, Cy; then, if we represent the differentiation with respect to x and y by d, and that with respect to c, and c, by d', we can obtain the singular solution by eliminating cı, cz, and Comme between the equations

U = 0, dU = 0, d'U = 0, dd'U = 0. (19) Let the given integral be

y = {cx? + co x + ' + cap, so that U = {4, 2? + C2 x + 0,? + ci- y = 0. Then dU = (C, « + c) dx dy = 0,

d'U = (5a + 2c7) dc, + (x + 2c,) dc, = 0,

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

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Substituting this in the preceding equation we have

4 (C – C, X) – x2 = 0. Between this equation and the first two we can eliminate ay and cu, and we find as the singular solution of the first integral

dy
) + ( x + 1) x - -y (1 + x>) = 0.

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(20) Let the integral be

Y = *Cira + C9 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

1 x dy) 2 dy
- + - + 3x

1- 4 xy = 0.
4 dx)

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

2:3 + 2y = 0.

Singular Solutions of Partial Differential Equations.

If U = 0 be a partial differential equation of the first order in x, y, and «, and if we put = p, = 9, the singular solution, if there be one, will be found by eliminating p and q between the three equations

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