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Exercises 5.2
Therefore, let’s try the substitution
z
(
t
)=
x
(
t
)+
y
(
t
). We want a function
z
(
t
) that satisFes
the two equations
z
0
(
t
)
−
z
(
t
−
3
e
−
2
t
and
z
0
(
t
)+2
z
(
t
)=3
e
t
,
(5.9)
simultaneously. We start by solving the Frst equation given in (5.9). This is a linear di±erential
equation with constant coeﬃcients which has the associated auxiliary equation
r
−
1=0
.
Hence, the solution to the corresponding homogeneous equation is
z
h
(
t
Ce
t
.
By the method of undetermined coeﬃcients, we see that a particular solution will have the
form
z
p
(
t
Ae
−
2
t
⇒
z
0
p
=
−
2
Ae
−
2
t
.
Substituting these expressions into the Frst di±erential equation given in (5.9) yields
z
0
p
(
t
)
−
z
p
(
t
−
2
Ae
−
2
t
−
Ae
−
2
t
=
−
3
Ae
−
2
t
=
−
3
e
−
2
t
⇒
A
=1
.
Thus, the Frst equation given in (5.9) has the general solution
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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