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The integral of this is
u + x + y + x = y* f {v (u – y), « (y – x)}.

Monge's Method
Let the partial differential equation be of the form

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equation

m? Pm + R = 0; and if from these two systems we can find two integrals U = a, V = b, then

V = f(U) is the first integral of the proposed equation ; and the integral of this is the complete integral of the proposed equation. It is generally more convenient (when possible) to find another first integral, of the from . . .

v'= f,(U'), and between these to eliminate p or q so as to obtain an equation involving only one differential coefficient, and which is therefore easily integrable.

Monge, Mémoires de l'Académie des Sciences, 1784, p. 118.

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the integral of which is

..(24- a) (- 9) + 2x = b = f (y - x), and therefore

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x + y + y From the first of (2) we find

y + x = ajo and substituting this in the equation just found, it becomes

dx dx 2x _ f (y x)

dy dva a This is a linear equation, and is therefore easily integrated. The result is

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aj where x + y is to be substituted for a, after integration.

(12) Let the equation be

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If we put p + q = a, this takes the form

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represents the integral of the proposed equation.

Poisson* has shewn how to obtain a particular integral of equations of the form

P = (rt - 52)" Q .................. (1) where P is a function of p, q, r, s, t, homogeneous with respect to the last three quantities, and Q is a function of x, y, x, and the differentials of z, which does not become infinite when rt - s = 0.

* Correspondance sur l'Ecole Polytechnique, Vol. 11. p. 410.

If we assume q= f(), we have

8 = rf'(p), t = 8f' () =r{f'(p)}? ;

and therefore rt – gể = 0. ............ (2) Hence the equation (1) is reduced to

P= 0; and on substituting in it the values of 9, 8, and t, the quantity go will divide out, as P is homogeneous in r, s, and t, and the equation is reduced to the form

F{p, f(p)f' (p)} = 0, which is an ordinary differential equation, and being integrated determines the form of f (p) involving an arbitrary constant. The partial differential equation

q = f (p) can always be integrated, and furnishes a value of x involving an arbitrary function and an arbitrary constant. This process comes to the same as finding what developable surfaces satisfy the equation (1).

(13) Let go? - t = rt - $?.
Assuming q= f (p) we find

gel {1 - [8"(p)]'} = 0, whence

ľ (p) = = 1; and therefore q= f (p) = + P + C, C being an arbitrary constant. On integrating this we find

%= Cx + (y + x) as a particular integral of the given equation.

(14) Let t + 2p8 + (p? - a) r = 0. .............. (1)

In this case Q = 0, and on putting 9 = f () we have, after dividing by r,

{f'(p)}" + 2 pf'(p) + p – a' = 0,; .........(2) from which

f' (p) + p = = a, and therefore

q+p ap = C........ .....

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