Page 312 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

So the solution is Notes
5
z = F 1 (y x ) F 2 (y x ) (2x y ) 3
6

Self Assessment

7. Solve

(D 2 3DD D 2 )z (x y )
8. Solve

(D 2 D 2 )z cos(mx ny )
Particular case when F(a, b) = 0

1 n 1
As (ax by ) = (ax by )
a
F ( ,D ) F ( , )
D
b
but if ( , ) 0F a b then R.H.S. becomes infinite and the above method fails.
Now consider the case
(bD aD )z = x r (ax by )

or bp aq = x r (ax by where ...(16)
),
z z
p = , q .
x y
Applying Lagrange’s method to (1) we get

dx dy dz
=
b a x r (ax by )
So one solution is

ax by = c, and the other solution is given by
dx dz
= r
b x ( )
c

x r 1
z = (ax by )
(r 1)b
This is the solution of the given differential equation (16).

1 r x r 1
Thus x (ax by ) = (ax by ) ...(17)
(bD aD ) ( b r 1)
Next consider

1
z = (ax by )
(bD aD ) n

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