Page 325 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes u
Put x e , y e
z z 1 z 1 z
= ,
x u x y y

2
x x z = z
x x x 2

2 z z 2 z
or x 2 2 x = 2
x x u
2 z z 2 z
2
and y 2 y =
y y 2
The equation (1) becomes
2
z z u 2
2 = e
2
The complementary function is
u
= 1 ( ) e 2 2 ( )
u
x
= 1 (log ) y 2 2 (log )
x
x
x
= ( ) y 2 ( )
1 2
1 u 2
P.I. = e
D (D 2)
1 u 2
= e
D (D 2)

e u 2 1 e u 2
= (1) .
2 (D 2 2) 2

1 2
= xy log y
2
xy 2
2
x
x
The solution is z = 1 ( ) y 2 ( ) log y
2
Aliter. yt q xy
The equation can be written as

q 1
q = x
y y
Solving,
1 1
dy dy
. q e y = xe y dy 1 ( )
y

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