Page 328 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 20: Higher Order Equations with Constant Coefficients and Monge’s Method

Self Assessment Notes

15. Solve

1 2 z 1 z 1 2 z 1 z
0
x 2 x 2 x 3 x y 2 y 2 y 3 y
16. Solve

2 z 2 z
x 2 y 2 xy
x 2 y 2

20.8 Monge’s Method

We shall usually take z as dependent and x, y as independent variables and throughout this
chapter we shall denote

z z 2 z 2 z 2 z
by ,p by ,q 2 by ,r by s, and by t.
x y x x y y 2

Monge’s Method of Solving the Equation

Rr Ss Tt = V ...(1)

where r, s, t have their usual meanings and R, S, T and V are functions of x, y, z, p and q.
We know

p p
dp = dx dy
x y
= r dx s dy

q q
and dq = dx dy
x y

= s dx + t dy.
Putting the values of r and t in (1),

dp s dy dq s dx
R . S s T . = V
dx dy

or R dp dy T dq dx Ss dxdy Rs dy 2 Ts dx 2 = V dx dy

or (R dp dy T dq dx V dx dy ) = s (R dy 2 S dx dy T dx 2 ) ...(2)

If some relation between x, y, z, p, q makes each of the bracketed expressions vanish, the relation
will satisfy (2); therefore

R dy 2 S dx dy T dx 2 = 0 ...(3)

Rdp dy T dq dx V dx dy = 0 ...(4)

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