Page 327 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
2 z 2 z 2 z z
3 4
Example 4: Solve: x 2 2 4xy 4y 3 2 6y x y .
x x y y y
x
Solution: As shown in the last example, if u log , log ,
y
z z z z
x = , y
x u y

2 z z 2 z 2 2 z z 2 z
x 2 x = 2 and y 2 2
x 2 x u y y y

z
Now y x =
t x u

3 z 2 z
or yx = .
x y u

With these substitution the equation takes the form
2 2 2
z z 4 z 4 4 z 6 z 3u 4
u 2 u u 2 = e .e

2 z 2 z 2 z z z
or 2 4 4 2 2 = e 3u 4 ...(1)
u u u

Denoting by D and by D in (1).
u

)
(D 2 4DD 4D 2 D 2D z = e 2u 4 .
2u 4
[(D 2D )(D 2D 1)]z = e
The complementary function is

= 1 ( 2u e u 2 ( 2 ).
u
2
2
= 1 (log x y ) x 2 (log x y )
2
2
= (x y ) x (x y )
1 3u 4
P.I. = e
(D 2D )(D 2D 1)

3 4
1 3u 4 x y
= e
( 5)( 6) 30
The solution is

3 4
x y
2
2
z = (x y ) x (x y ) .
30
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