1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.2 Nonhomogeneous Linear Equations 21

or, after canceling 5ve^5 tfrom both sides and simplifying, we find

dv

dt=e

− 5 tt.

This equation is integrated once (by parts) to find

v(t)=

(

−t

5

−^1

25

)

e−^5 t.

From here, we obtainup(t)=v(t)·e^5 t=−

( 1

5 t+

1

25

)

.

2. Second-order equations

To find a particular solution of the nonhomogeneous second-order equation

d^2 u

dt^2 +k(t)

du

dt+p(t)u=f(t), (9)

we need two independent solutions,u 1 (t)andu 2 (t),ofthecorrespondingho-
mogeneous equation

d^2 u

dt^2

+k(t)du

dt

+p(t)u= 0. (10)

Then we assume that our particular solution has the form

up(t)=v 1 (t)u 1 (t)+v 2 (t)u 2 (t), (11)

wherev 1 andv 2 are functions to be found. If we simply insertupin this form
intoEq.(9),weobtainonecomplicatedsecond-orderequationintwoun-
known functions. However, if we impose the extra requirement that

dv 1

dt

u 1 +dv^2

dt

u 2 = 0 , (12)

then we find that

u′p=v 1 ′u 1 +v 2 ′u 2 +v 1 u′ 1 +v 2 u′ 2 =v 1 u′ 1 +v 2 u′ 2 , (13)

u′′p=v 1 ′u′ 1 +v 2 ′u′ 2 +v 1 u′′ 1 +v 2 u′′ 2 , (14)

and the equation that results from substituting Eq. (11) into Eq. (9) becomes

v 1 ′u′ 1 +v′ 2 u′ 2 +v 1 (u′′ 1 +k(t)u′ 1 +p(t)u 1 )+v 2 (u′′ 2 +k(t)u′ 2 +p(t)u 2 )=f(t).

This simplifies further: The multipliers ofv 1 andv 2 are both 0, becauseu 1 and
u 2 satisfy the homogeneous Eq. (10).