Page 326 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
q x
= dy ( )
x
y y 1
x
q = xy log y y 1 ( )
z
or = xy log y y 1 ( )
x
y
y 2
x
x
z = x y log ydy 1 ( ). 2 ( )
2
y 2 y 2 1
2
x
x
= x log y dy y f ( ) F ( )
2 2 y
xy 2 xy 2
2
x
x
z = log y y f ( ) F ( )
2 4
is the required solution.
2 z z z z
Example 3: Solve: x 2 y 2 y x 0
x 2 y 2 x x
Solution: Assume u log , log . Then
x
y
z z 1
=
x u x
z z
or x = , so that x ...(1)
x x x x
z 2 z z 2 z
x x = x 2 x [from (1)]
x x x 2 x u 2
Similarly
2 z z
y 2 2 y = z .
y y y 2

The given equation reduces to
2 2
z z
u 2 2 = 0,
for which
z = (u ) ( ) u

y
x
= [logx log ] [log y log ]
y
= (log xy ) log
x

y
= f 1 (xy ) f 2
x

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