1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.2 Nonhomogeneous Linear Equations 17

Now, equating coefficients of like terms gives these two equations for the coef-
ficients:

6 A 0 = 1 (coefficient ofte−t),

6 A 1 − 2 A 0 = 0 (coefficient ofe−t).

ThesewesolveeasilytofindA 0 = 1 /6,A 1 = 1 /18. Finally, a particular solution
is

up(t)=

( 1

6

t+^1

18

)

e−t. 

A trial solution from Table 4 will not work if it contains any term that is a so-
lution of the homogeneous differential equation. In that case, the trial solution
has to be revised by the following rule.

Revision Rule.Multiply by the lowest positive integral power of t such that no
term in the trial solution satisfies the corresponding homogeneous equation.

Example.
Table 4 suggests the trial solutionup(t)=(A 0 t+A 1 )e−tfor the differential
equation

d^2 u

dt^2 −u=te

−t.

However, we know that the solution of the corresponding homogeneous equa-
tion,u′′−u=0, is

uc(t)=c 1 et+c 2 e−t.

The trial solution contains a term(A 1 e−t)that is a solution of the homoge-
neous equation. Multiplying the trial solution byteliminates the problem.
Thus,thetrialsolutionis

up(t)=t(A 0 t+A 1 )e−t=

(

A 0 t^2 +A 1 t

)

e−t.

Similarly, the trial solution for the differential equation

d^2 u

dt^2 +^2

du

dt+u=te

−t

has to be revised. The solution of the corresponding homogeneous equation
isuc(t)=c 1 e−t+c 2 te−t. The trial solution from the table has to be multiplied
byt^2 to eliminate solutions of the homogeneous equation.

Example: Forced Vibrations.
The displacementu(t)of a mass in a mass–spring–damper system, starting