Download When solving a fixed-constant linear ordinary differential equation

When solving a fixed-constant linear ordinary differential equation where the
characteristic equation has repeated roots, why do we get the next independent solution in
the form of x n e mx ? Show this through an example.
Let’s suppose we want to solve the ordinary differential equation
d2y
dy
+ 6 + 9y = 0
2
dx
dx
The characteristic equation is
m 2 + 6m1 + 9m 0 = 0
m 2 + 6m + 9 = 0
(m + 3) 2 = 0
The solution to the characteristic equation is
m = −3,−3
and the homogeneous part of the solution to the ordinary differential equation is
y H = K 1e −3 x + K 2 xe −3 x
Since the forcing function is zero, the particular part of the solution to the ordinary differential
equation is
yP = 0
The complete solution to the ordinary differential equation is
y = yH + yP
= K1e −3 x + K 2 xe −3 x
But why is K 2 xe −3 x part of the solution? See the answer below.
d2y
dy
+ 6 + 9y = 0
2
dx
dx
can be rewritten as
d
 d

 + 3  + 3  y = 0
 dx
 dx

(1)

d
 + 3 y = z

 dx
(2)
Let
Then equation (1) can be written as

d
 + 3 z = 0

 dx
(3)
Let
z = K 2 e m2 x
be a possible solution form of equation (3), then

d
m x
 + 3 K 2 e 2 = 0
dx


(
)
K 2 m 2 e m2 x + 3 K 2 e m2 x = 0
K 2 (m2 + 3)e m2 x = 0
Then
m2 = −3
is a solution which is nontrivial.
So
z = K 2 e −3 x
From equation (2)

d
 + 3 y = z

 dx
d

−3 x
 + 3 y = K 2 e
 dx

dy
+ 3 y = K 2 e −3 x
dx
Multiply both sides by e 3 x

 dy
e 3 x  + 3 y  = K 2 e −3 x e 3 x

 dx
e3x
dy
+ 3e 3 x y = K 2
dx
e3x
d
d
( y ) + y (e 3 x ) = K 2
dx
dx
By chain rule
d 3x
e y = K2
dx
(
)
Integrating both sides with respect to x , we get
e 3 x y = K 2 x + K1
y = K 2 xe −3 x + K 1e −3 x