Nonhomogeneous Linear Differential Equations with Constant Coefficients - (3.4)
Method of Undetermined Coefficients
Consider an nth-order nonhomogeneous linear differential equation with constant coefficients:
LŸy gŸx where LŸy a n y Ÿn a n"1 y Ÿn"1 . . . a 1 y U a 0 y.
We know the general solution of this differential equation is:
y yh yp
where
y h C 1 y 1 . . . C n y n
is the general solution of LŸy 0 and y p is a particular solution of LŸy gŸx , respectively. We know how
to find y h from Section 3.3. In this section, we will study a method called “The Method of Undetermined
Coefficients” to find y p .
Notice that if gŸx g 1 Ÿx g 2 Ÿx and y p 1 and y p 2 are particular solutions of LŸy g 1 Ÿx and
LŸy g 2 Ÿx , respectively, then y p y p 1 y p 2 is a particular solution of LŸy gŸx . So if gŸx is a sum of k
functions g i Ÿx U s, we may solve equations LŸy g i Ÿx for i 1, T , k one by one.
The Method of Undetermined Coefficients:
Let y h C 1 y 1 . . . C n y n be the general solution of the differential equation:
LŸy 0.
Find y p , a particular solution of the differential equation:
LŸy gŸx .
Observe that the following are possible types of functions for y Ui s :
type of gŸx polynomial
examples
1 x, x 2 "
1
3
e"
1
2
x3
exponential function
e 2x ,
2x
sine and cosine
cosŸ=x , sinŸ2x combinations of above functions
xe 2x , x sinŸ3x , Ÿx 2 " 1 e x cosŸ2x This method is designed to solve y p when gŸx is one of above functions.
1. The type of gŸx is different from any of y Ui s. The solution y p can be chosen as follows.
type of gŸx choice of y p
b k x k b k"1 x k"1 . . . b 1 x b 0
A k x k A k"1 x k"1 . . . A 1 x A 0
e ax
Ae ax
cosŸ*x or sinŸ*x )x
)x
A cosŸ*x B sinŸ*x e cosŸ*x or e sinŸ*x Ae )x cosŸ*x Be )x sinŸ*x Ÿb k x k b k"1 x k"1 . . . b 1 x b 0 e )x
ŸA k x k A k"1 x k"1 . . . A 1 x A 0 e )x
Ÿb k x k b k"1 x k"1 . . . b 1 x b 0 sinŸ*x or
ŸA k x k A k"1 x k"1 . . . A 1 x A 0 sinŸ*x Ÿb k x k b k"1 x k"1 . . . b 1 x b 0 cosŸ*x ŸB k x k B k"1 x k"1 . . . B 1 x B 0 cosŸ*x 2. The type of gŸx is the same as one of y Ui s. The solution y p x h y p where y p is chosen from above table
and the positive integer h is chosen so that x h y p is different from any of y Ui s.
1
Constants A Ui s and B Ui s are determined so that y p is a particular solution.
Example Let y h be the general solution of LŸy 0 where
y h C 1 e 2x C 2 xe 2x C 3 cosŸ2x C 4 sinŸ2x C 5 e "2x C 6 e "x sinŸ=x C 7 e "x cosŸ=x C 8 .
Give the form of a particular solution y p of the differential equation:
LŸy 2 x 2 e 3x 4e 2x cosŸ=x " 2x sinŸ4x " 2e x cosŸ2x " cosŸ2x .
Consider gŸx g 1 Ÿx g 2 Ÿx g 3 Ÿx g 4 Ÿx g 5 Ÿx g 6 Ÿx g 7 Ÿx . Choose y p i :
i g i Ÿx y pi
1 2 x2
ŸA 2 x 2 A 1 x A 0 x
2 e 3x
Be 3x
3 4e 2x
Ce 2x Ÿx 2 4 cosŸ=x D 1 cosŸ=x D 2 sinŸ=x 5 "2x sinŸ4x ŸE 1 x E 2 sinŸ4x ŸE 3 x E 4 cosŸ4x 6 "2e x cosŸ2x F 1 e x cosŸ2x F 2 e x sinŸ2x 7 " cosŸ2x ŸG 1 cosŸ2x G 2 sinŸ2x x
y p y p 1 . . . y P 7
ŸA 2 x 2 A 1 x A 0 x Be 3x Ce 2x Ÿx 2 D 1 cosŸ=x D 2 sinŸ=x ŸE 1 x E 2 sinŸ4x ŸE 3 x E 4 cosŸ4x F 1 e x cosŸ2x F 2 e x sinŸ2x ŸG 1 cosŸ2x G 2 sinŸ2x x
Example Solve y UU " 2y U " 3y 4x " 5 6xe 2x .
1. Solve y h from the equation: y UU " 2y U " 3y 0.
PŸm m 2 " 2m " 3 Ÿm " 3 Ÿm 1 0, m 3, m "1
The general solution of LŸy 0 :
y h c 1 e 3x c 2 e "x
2. Solve y p 1 from y UU " 2y U " 3y 4x " 5.
Let y p 1 Ax B. Then
y Up 1 A, y UUp 1 0.
y UU " 2y U " 3y 4x " 5 « 0 " 2A " 3ŸAx B 4x " 5 « "3Ax Ÿ"2A " 3B 4x " 5
coefficients of x : "3A 4, A " 43
constants: " 2A " 3B "5, B 1
3
5"2 "
4
3
23
9
, y p 1 " 4 x 23
9
3
3. Solve y p 2 from y UU " 2y U " 3y 6xe 2x .
Let y p 2 ŸAx B e 2x . Then
y Up 2 Ae 2x 2ŸAx B e 2x Ÿ2Ax A 2B e 2x ,
y UUp 2 Ÿ2A e 2x 2Ÿ2Ax A 2B e 2x Ÿ4Ax 4A 4B e 2x
y UU " 2y U " 3y 6xe 2x « Ÿ4Ax 4A 4B e 2x " 2Ÿ2Ax A 2B e 2x " 3ŸAx B e 2x 6xe 2x
Dropping e 2x from both sides of the equation, we have polynomials on both sides of the equation:
Ÿ4A " 4A " 3A x Ÿ4A 4B " 2A " 2B " 3B 6x « "3Ax Ÿ2A " B 6x
2
coefficients of x : "3A 6, A "2
constants: 2A " B 0, B 2A "4
, y p 2 Ÿ"2x " 4 e 2x "2Ÿx 2 e 2x
4. the general solution of LŸy gŸx :
y y h y p 1 y p 2 c 1 e 3x c 2 e "x " 4 x 23 " 2Ÿx 2 e 2x
9
3
Example Solve y UUU " 5y UU 4y U 8e x 4x.
1. Solve y h from y UUU " 5y UU 4y U 0.
PŸm m 3 " 5m 2 4m mŸm " 4 Ÿm " 1 0, m 0, m 1, m 4.
The general solution of LŸy 0 :
y h c 1 c 2 e x c 2 e 4x .
2. Solve y p 1 from y UUU " 5y UU 4y U 8e x .
Let y p 1 ŸAe x x Axe x . Then
y U AŸe x xe x AŸ1 x e x , y UU AŸe x Ÿ1 x e x AŸ2 x e x
y UUU AŸe x Ÿ2 x e x AŸ3 x e x
y UUU " 5y UU 4y U 8e x « AŸ3 x e x " 5AŸ2 x e x 4AŸ1 x e x 8e x
Drop the factor e x from both sides of the equation, we have polynomials in x on both sides of the equation:
ŸA " 5A 4A x Ÿ3A " 10A 4A 8 « "3A 8, A " 8 , y p 1 " 8 xe x
3
3
UUU
UU
U
3. Solve y p 2 from y " 5y 4y 4x
Let y p 2 ŸAx B x Ax 2 Bx. Then
y Up 2 2Ax B, y UUp 2 2A, y UUU
p 3 0.
y UUU " 5y UU 4y U 4x « 0 " 5Ÿ2A 4Ÿ2Ax B 4x « 8Ax Ÿ"10A 4B 4x
coefficients of x : 8A 4, A constants: " 10A 4B 0, B 1
2
10
4
A
5
4
, y p2 1 x 2 5 x
2
4
4. The general solution of LŸy gŸx :
y y c y p 1 y p 2 c 1 c 2 e x c 2 e 4x " 8 xe x 1 x 2 5 x
3
2
4
UU
Example Solve y 9y 2 cosŸ2x 0
if 0 t x =
2 , yŸ0 0, y U
Ÿ0 0.
=
if x u
2
1. Solve y h from y UU 9y 0.
PŸm m 2 9 0, m oi3.
The general solution of LŸy 0 :
y h c 1 cosŸ3x c 2 sinŸ3x 2. Solve y p from y UU 9y 2 cosŸ2x .
Let y p A cosŸ2x B sinŸ2x . Then
y Up "2A sinŸ2x 2B cosŸ2x , y UUp 2 "4A cosŸ2x " 4B sinŸ2x y UU 9y 2 cosŸ2x « "4A cosŸ2x " 4B sinŸ2x 9ŸA cosŸ2x B sinŸ2x 2 cosŸ2x 3
Ÿ"4A 9A cosŸ2x Ÿ"4B 9B sinŸ2x 2 cosŸ2x coefficients of cosŸ2x : 5A 2, A 2
5
coefficients of sinŸ2x : 5B 0, B 0
, y p 2 cosŸ2x 5
3. The general solution of LŸy gŸx :
c 1 cosŸ3x c 2 sinŸ3x y yh yp 4. Solve the initial value problem:
For 0 t x = ,
2
c 3 cosŸ3x c 4 sinŸ3x 2
5
cosŸ2x 0 t x =
2
=
xu
2
y U "3c 1 sinŸ3x 3c 2 cosŸ3x " 4 sinŸ2x 5
yŸ0 c 1 2
5
0, c 1 " 25
,
y U Ÿ0 3c 2 0, c 2 0
y " 2 cosŸ3x 2 cosŸ2x .
5
5
For x u = , the initial conditions are:
2
y 6 sinŸ3x " 4 sinŸ2x ,
5
5
U
y =
2
" 25
yU =
2
" 65
Then
y U "3c 3 sinŸ3x 3c 4 cosŸ3x y =
2
yU =
2
0 " c 4 " 25 , c 4 6
5
3c 3 " , c 3 "
2
5
2
5
, y " 2 cosŸ3x 2 sinŸ3x 5
5
The solution of the initial value problem:
y yh yp 4
" 25 cosŸ3x 2
5
" 25 cosŸ3x 2
5
cosŸ2x 0 t x =
2
=
sinŸ3x x u
2
2
0.4
1
0.2
0
0.5
1
1.5
x
2
2.5
3
0
2
4
x
-0.2
-1
-0.4
-2
fŸx 5
2 cosŸ2x if 0 t x 0
if x u
=
2
=
2
y
6
8
10