Page 133 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes ax
V e ax = Qe dx c

V = e ax Qe ax dx ce ax .

Now c can be taken zero, for we want only a particular solution.

Hence V = e ax Qe ax dx .

1
or Q = e ax Q e ax dx .
(D )
We are now in a position to evaluate

1
D
f
{ ( )} Q
Let on factorization

f ( ) = (D 1 )(D 2 ) (D n )
D

Then (D 1 )(D 2 ) (D n )y = Q
It follows that
1
(D 1 )(D 2 ) (D n )y = (D 1 ) Q

= e 1 x e 1 x Qdx
Therefore

1

(D 3 ) (D n )y = (D 2 ) e 1 x e 1 x Qdx

or (D 3 ) (D n )y = e 2 x e ( 1 2 )x e 1 x Qdx
and so on.
Hence, we get generally

y = e n x e ( n 1 a n )x  e ( 1 2 )x e 1 x Qdx dx .

This is the required particular integral.
Note: In case f(D) fails to give real linear factors, we may use imaginary factors and use the above
method and finally put the result in a real form.

1
Let be capable of resolving into partial fractions. Thus
f ( )
D
1 A A A
= 1 2  n
f ( ) D D D
D
1 2 n
Now, particular integral
1 A A A
= Q 1 Q 2 Q ... n Q .
D
f ( ) D D D
1 2 n

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