Page 345 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes and from (2) and (1),

1 p x 1 1
q = F F F ( ) [from (2 )]
y y y y y

1
= . ( )  From (1) and (3), ...(4)
F
y
Now

dz = p dx + q dy

= ( ) x dx F ( )dy [from (3) and (4)]

x 2
z = x F ( )y k
2

x 2 1
= x F y ( )
2 y y

x 2
or z = x F ( )
2 y

Example 6: Solve:
5r 6s 3t 2(rt s 2 ) 3 = 0 ...(1)

Solution: Comparing it with

Rr Ss Tt U (rt s 2 ) = V

We have R = 5, S 6, T 3, U 2, V 3
The -quadratic will be

2 2
(UV RT ) SU U = 0
or 9 2 12 4 = 0

or (3 2) 2 = 0

2 2
= , .
2 3 3
The intermediate integral will be
U dy 1 T dx 1 . U dp = 0

and 2 R dy U dx 2 . U dq = 0

or 3 dy 3 dx 2dp = 0 and 5dy 3 dx 2 dq 0.

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