Page 307 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Hence the solution is
(z x F (y mx y mx ) = 0
),
or z x ( F y mx ) F 1 (y mx )

so z = xF (y mx ) F 1 (y mx ) ...(10)

In general, the solution of
r
(D mD ) z = 0
is z = F 1 (y mx ) xF 2 (y mx ) ... x r 1 F r (y mx ) ...(11)

Example 1: Solve
4 4 4 4
z 2 z 2 z z
3
x 4 x y x y 3 y 2 = 0
The auxiliary equation is
m 4 2m 3 2m 1 = 0

m
m 4 1 2 (m 2 1) = 0
m
(m 2 1)(m 2 1) 2 (m 2 1) = 0
(m 2 1)(m 1) 2 = 0 (m 1)(m 1) 3

So the roots are 1, 1, 1, 1
Hence the solution is
2
z = F (y x ) xF (y x ) x F (y x ) F (y x )
1 2 3 4
Example 2: Solve

(25D 2 40DD 16D )z = 0
The auxiliary equation is

25m 2 40m 16 = 0
(5m 4) 2 = 0

4
The roots are m , 4 5 are repeated roots so the solution is
5
x
z = F 1 (5y 4 ) x F 2 (5y 4 )
x
Self Assessment

3. Solve
3 3 3
z 4 z 4 z 0
2
x 3 x y x y 2

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