Page 324 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

The solution is Notes

x 2
x
x
y
x
u = 1 ( ) (y 2 ) x (y 2 ) sin(y 2 ).
4
Self Assessment
2 z 2 z z z
2
13. Solve 2 a 2 2ab 2a b 0
x y x y
2 2
z z z
y
14. Solve 2 z cos (x 2 )
x x y y
20.7 Equation Reducible to Homogeneous Linear Form

An equation in which the coefficient of a differential coefficient of any order is a constant
multiple of the variables of the same degree may be transformed into one having constant
coefficients. The method is explained with the help of the following equations.

Example 1: Solve
2 z 2 z 2 z
x 2 2 2xy y 2 2 = 0
x x x y

y
x
Solution: Assume, u log , v log , also denoting by D and by D , the given equation
u V
reduces to
[ (D 1) 2DD D (D 1)]z = 0
D
or (D D )(D D 1)z = 0

Hence the solution is

z = ( ) u e u ( ) u
1 2
x
x
= 1 (log y log ) 2 (log y log )
y y
= 1 log x 2 log
x x
y y
= 1 x 2
x x

Example 2: Solve: yt q xy .
Solution: The equation can be written as

2 z z
y 2 2 y = xy 2 ...(1)
y y

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