Page 203 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes This can be rearranged to show that
2
2
d u du d u du du dc du dc
c 1 2 1 a 1 1 a u c 2 2 2 a 1 2 a u 1 1 2 2 f
0 2
0 1
dx dx dx dx dx dx dx dx
Since u and u are solutions of the homogeneous equation, the first two terms are zero, which
1 2
gives us
du dc du dc
1 1 2 2 f ...(5)
dx dx dx dx
We now have two simultaneous equations (4) and (5), for c = dc /dx and c dc /dx which can
,
1 1 2 2
be written in matrix form as
u 1 u 2 c ' 1 0
u ' 1 u ' 2 c ' 2 f

These can easily be solved to give
fu fu
c 2 ,c 1 ,
1 2
W W
where

u 1 u 2
W u u u u
1 2
2 1
u 1 u 2
is called the Wronskian. These expansions can be integrated to give
s
u
s
s
x f ( ) ( ) x f ( ) ( )
u
s
c 1 2 ds  , A c 2 1 ds . B
s
s
W ( ) W ( )
We can now write down the solution of the entire problem as
u
s
u
s
x f ( ) ( ) x f ( ) ( )
s
s

x
y ( ) u 1 ( ) 2 ds u 2 ( ) 1 ds Au 1 ( ) Bu 2 ( )
x
x
x
x
W ( ) W ( )
s
s
The particular integral is therefore
x u ( ) ( ) u ( ) ( )
s
u
x
u
s
x
s
x
y ( ) f ( ) 1 2 1 2 ds ...(6)
s
W ( )
This is called the variation of parameters formula.
Example: Consider the equation
2
d y
y x sinx
dx 2
The homogeneous form of this equation has constant coefficients, with solutions
u (x) = cos x, u (x) = sin x
1 2
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