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du

(9) Let

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du

du
+ (u + x) + (4 + y)
do
dy

dz
The integral of this is

u + x + y + z = y'f {x(u y), « (y – x)}.

Monge's Method.

Let the partial differential equation be of the form

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ď x dx dy

R,

}... (1)
....

dx

where p =

and a

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+ P + Q dx2

dy

dx dz where P, Q, R are functions of x, y, x,

dx' dy Then if we form the system of equations

dy - mdx = 0
mdp + Qdq - Rmdx = 0
dy m'dx = 0

(2)
m'dp + Qdq - Rm'dx = 0
dx

and m, m' are the roots of the d.

dy 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

'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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(10) Let q?

dx?

2 p q

dx dy

+ p?

dy

= 0.

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From the second, since dx = pdx +qdy, we have

0,

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or = a.

From the third we have

pdq - qdp = 0;

р
whence 0 () = C,
a

From
since z = constant, and therefore p (p) = constant.
the equation p - cq = 0, we easily obtain

= f (x + cy) = f {x + y (x)}, which is the required integral.

d'% da?

dy' x + y The auxiliary cquations are, dy - dx = 0, dp - dq +

0 ...... (1) x + y

d?z

(11) Let

+

= 0.

4p

dx =

dy + dx = 0, dp + dq

4p

dx + y

= 0 ......

... (2)

From the first of (1) we find

4 pdx Y X = C, and therefore dp - dq +

= 0.

2y - a If we subtract from this last the equation P

2pdy 2dx 2qdy (dy-dr) =

+ 2y - a

2y - a 2 y -a 2y a (as pdx = dz - gdy) we have

(2y a) (dp - dq) + 2 (p 9) dy + 2dz

0;

1 2 y

2 y

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the integral of which is

(2y - a) (p - 9) + 2x = b = f (y – X), and therefore

2x

f(y x)
P - 9+
at + y

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

y + x = a je and substituting this in the equation just found, it becomes dx dx 2x

f(y x)
dy dx
ai

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

x

+ 6+ (x + y),

aj where x + y is to be substituted for a, after integration. (12) Let the equation be dx

ď % (1 + pq +98) + (qo p) (1 + pq + p) 0. d.ro dxdy

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

đ , (1 + qa) + (9-p) a (1 + pa)

= 0.
da

dy
The equation for determining m is
(1 + qa) m? (q p) am – (1 + pa) = 0;

1 + pa which gives

m = 1,

1 +qa We have therefore to integrate the two systems, dy - dx = 0; dp (1 + qa) - dq (1 + pa) = 0 ... (1), dy (1 + qa) + dv (1 + pa) = 0; dp + dq = 0 ... (2). The second equation of (2) gives p + 9 = b or a = b. The first equation of (2), when put under the form

dx + dy ta (pdx + qdy) = 0,

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đ , dxdy

m =

ada

2

ܪ

2 ta

gives

x + y + (P + 9) =

= a; therefore

x + y + (p +9)= 0 (p +9). The first equation of (1) gives y - x = a, and putting p - 9 = ß, we have

P } (a + b), q = à la B),

dp = 1 (da + dB), dq = } (da - dB), and therefore the second equation of (1) may be put under the form 13

whence B = 1; (2 + a?)?, B and therefore

P-q = y (x – y) {2 + (P + 9)?}!". This first integral will enable us to determine the second integral. Putting p +9

Putting p + q = a, p - 9 = ß, we have dx = (a +B)dx + (a-B)dy = }a (dx + d y) + B(dx dy); or, putting for B its value y (x y) (? + a?)",

di = 1 a (dx + dy) + (1 x dy) y (x y) (2 + a?)!.

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

te + (a) = 1 a (x + y) + y (x - y) (2 + a?)}; which, combined with

a

$'(a) = } (x + y) + y (.r y)

(2 + a“) represents the integral of the proposed equation.

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

P = (rt s?)" 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 2, which does not become infinite when - s2 = 0.

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

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

8 = rf'(p), t = sf'(o) = r {f'(p)}”;

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

P=0; and on substituting in it the values of q, 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

9 = f(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). (13) Let

pol-tert-8. Assuming 9 = f(p) we find

go? {1-[8'(p)]} whence

'

f (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 + + 2p8 + (p’ – a) r = 0. ............... (1)

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

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

f'(p) + p = a, and therefore 9 + £ p #ap = C....................(3)

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