Page 161 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
1 1
or c 2 = 0
x y x y
x y x y c
(x 2 y 2 ) 2 = 0

2y c 2 (x 2 y 2 ) = 0

2y
So c = 2 2
2 y x
So complete solution is

x y zy
c
c
( , ) = 0 , 2 2 0 ...(6)
1
2
z y x
Example 3: Solve
dx dy dz
= ...(1)
xy y 2 xyz 2x 2
Solution:

From the first two members
dx dy
= 2
xy y

dx dy
= ...(2)
x y
Integrating (2) we have
log x = log y logc 1

or x = c y ...(3)
1
From the second and third member of (1) we have

dy dz
y 2 = xyz 2x 2 ...(4)
Putting the value of x from (3) we have from (4)

dy = dz
2 2
y 2 [zc y 2 2c y ]
1
1
dz
or dy = ...(5)
(c z 2c 1 2 )
1
Integrating (5) we have

dz c
dy = 2
c
c (z 2 ) c
1 1 1

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