Page 129 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes This consequently means that if two roots of the auxiliary equation are equal, the general
solution of (i) will be

y = (c 1 c 2 ) x e m 1 x c e m 2 x ... c e m n x .
3
n
In general, if r roots of the auxiliary equation P m n P m n 1 ... P n 0 are equal to m say, the
0
1
1
general solution of (i) will be
y = (c 1 c x c x 2 ... c x r 1 )e m 1 x c r 1 e m r 1 x ... c e m n x .
2
r
3
n
Auxiliary Equation having Complex Roots
If some of the roots of auxiliary equation are complex, then we shall follow the procedure as
given below:

Let i be the roots of the auxiliary equation; then the corresponding part shall become
= c e ( i ) c e ( i )x
1
2
x
= c e x e i x c e e i x
1
2
c
= e ax ( cos x ic 1 sin x ) e ax (c 2 cos x ic 2 sin x )
1
= e ax [(c 1 c 2 )cos x (ic 1 ic 2 )sin x ]

A
= e ax [ cos x b sin x ],
where A and B are arbitrary constants.
Therefore the solution is
y = e ax ( cos x c 2 sin x ) c e m 2 x ... c e m n x
c
n
3
1
)
Example 1: The expression e ax (A cos x b sin x can be also written as
c e ax cos( x c 2 ) or c e ax sin( x c 2 ),
1
1
Example 2: if the auxiliary equation has two equal pairs of complex roots, say i
occurring twice, then the portion of the solution corresponding to these roots, is

e x (c 1 c x )cos x (c 3 c x )sin x
4
2
Example 3: If the auxiliary equation has the roots as , then the portion of the
solution corresponding to these roots is
c e ax cos h x c 2 or c e ax sin h x c 2
1
1
Solution of equations of the form

n
d y d n 1 y
P 0 P 1 ... P y 0.
n
dx n dx n 1

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