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u + x + y + z = y2ƒ {x (u − y), x (y − x)}.

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where P, Q, R are functions of x, y, z,

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and if from these two systems we can find two integrals Ua, 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 por 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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From the second, since de pdx+qdy, we have

dz

=

= 0, or = a.

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since

constant, and therefore

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and therefore (~) = constant.

the equation pcq = 0, we easily obtain

≈ = f(x + cy) = ƒ {x + y $(≈)},

which is the required integral.

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(2ya) (dp-dq) + 2 (p − q) dy + 2dx = 0;

From

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and substituting this in the equation just found, it becomes

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

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

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

da2 + (q − p) a

dx dy

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(1 + pa)

= 0.

dy

is

(1 + pa) = 0;

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dp (1+qa) - dq (1 + pa) = 0 (1),

...

dy dx = 0;
dy (1+qa) + dx (1 + pa) = 0; dp + dq=0... (2).

The second equation of (2) gives p + q = b or a = b.
The first equation of (2), when put under the form
dx + dy + a (pdx + qdy) = 0,

gives

therefore

x + y + (p + q) ≈ = a;

x + y + (p + q) ≈ = ¢ (p + q).

The first equation of (1) gives y p-q=B, we have

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x =

a, and putting

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dp = } (da + dB), dq = } (da – dB),

and therefore the second equation of (1) may be put under the form

αβ ada whence ẞb, (2 + a2)3,

В

and therefore

==

2+ a

2

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This first integral will enable us to determine the second integral. Putting p + q = a, pq = ẞ, we have

2=

ß,

: = 1 (a + ẞ) d x + 1 (a− ẞ) dy = ↓ a (dx+dy) + 1⁄2 ß (dx−dy);

or, putting for ẞ its value (x − y) (2 + a2)3,

dz = 1⁄2 a (dx + dy) + 1 (dx − dy) f (x − y) (2 + a2)1.

This is integrable if we suppose a to be constant, and gives

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≈ + 4 (a) = ¦ a (x + y) + √, (x − y) (2 + a2)1; which, combined with

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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.

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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, ≈, and the differentials of x, which does not become infinite when rt - s2 = 0.

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

If we assume q=f(p), we have

s = rf' (p), t = sf'(p) = r {ƒ′(p)}2 ;

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

F{p, f(p)ƒ' (p)} = 0,

which is an ordinary differential equation, and being integrated determines the form of ƒ (p) involving an arbitrary constant. The partial differential equation

q = ƒ (p)

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

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C being an arbitrary constant. On integrating this we find

≈ = Cx ± 4 (y ± x)

as a particular integral of the given equation.

(14) Lett+2ps + (p2 − a3) r = 0. ...............(1)

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

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