1.1 Basic Concepts and Formulae
19
First step: Write down the corresponding auxiliary equation
D
n
+ p 1 D
n−1
+ p 2 D
n−2
+ · · · + p n = 0
Second step: Solve completely the auxiliary equation.
Third step: From the roots of the auxiliary equation, write down the corresponding particular solutions of the differential equation as follows
Auxiliary equation
Differential equation
(a) Each distinct real root r 1
Gives a particular solution e
r 1 x
(b) Each distinct pair of imaginary
roots a ± bi
Gives two particular solutions
e
ax cos bx, e
ax sin bx
(c) A multiple root occurring s times Gives s particular solutions obtained by
multiplying the particular solutions
(a) or (b) by 1, x, x
2
, . . . , x
n−1
Fourth step: Multiple each of the n independent solutions by an arbitrary constant
and add the results. This gives the complete solution.
Type II
(I )
d
n y
dx n + P 1
d
n−1 y
dx n−1 + P 2
d
n−2 y
dx n−2 + · · · + P n y = X
where X is a function of x alone, or constant, and P 1 , P 2 , . . . P n are constants.
When X = 0, (I ) reduces to (A) Type I.
(J )
d
n y
dx n + P 1
d
n−1 y
dx n−1 + P 2
d
n−2 y
dx n−2 + · · · + P n y = 0
The complete solution of (J ) is called the complementary function of (I ).
Let u be the complete solution of (J ), i.e. the complementary function of (I ), and
v any particular solution of (I ). Then
d
n
v
dx n + P 1
d
n−1
v
dx n−1 + P 2
d
n−2
v
dx n−2 + · · · + P n v = X
and
d
n u
dx n + P 1
d
n−1 u
dx n−1 + P 2
d
n−2 u
dx n−2 + · · · + P n u = 0
Adding we get
d
n (u + v)
dx n
+ P 1
d
n−1 (u + v)
dx n−1
+ P 2
d
n−2 (u + v)
dx n−2
+ · · · + P n (u + v) = X
showing that u + v is a solution of I .