Page 337 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 20.9 M o n g e’s Method of Integrating Rr + Ss + Tt + U (rt - s ) = V
2
R, S, T, U are functions of x, y, z, p, q.
As before put
r = (dp s dy )/dx

and t = (dq s dx )/dy .
The equation reduces to
Rdpdy T dqdx U dpdq V dxdy ( s Rdy 2 Sdxdy T dx 2 U dpdx V dp dy ) = 0

or N Ms = 0.
So far, we used to factorise M, but on account of the presence of U dx dp + V dq dy, the factors are
not possible; so let us try to factorise M + N, where is some multiplier to be determined later.

Now N + M = (Rdpdy T dqdx U dpdq V dxdy )
(Rdy 2 Sdxdy T dx 2 U dpdx V dqdy )

= Rdy 2 T dx 2 (S V )dxdy U dpdx U dqdy Rdpdy T dqdx U dpdq .
Let the factors of the above be

dy dx dp and dy dx dq .
2
2
Equating coefficient of dy , dx , dp dq in the product,
= R , , T U .
Now if we take
k
= R , 1, kT , (1/ ), mU , /m
equating the coefficients of the other five terms.
kT R /k = (S V ). ...(1)

R /m = U, ...(2)
kT /m = T, ...(3)
mU = R, ...(4)
mU/k = U. ...(5)
From (5), m = k and this satisfies (3).

R
From (2) and (3), m R /U k . on putting k
U
From (1),

2 2
(RT UV ) US U = 0 ...(6)
The first step in practical working is to form the equation (6) in and to determine the two roots
and of this equation.
1 2
So if is a root of (6), factorised M + N is
1
RT U U
R dy 1 dx 1 R dp dy dx dq
U 1 R R

330 LOVELY PROFESSIONAL UNIVERSITY