Exercises 5.2
19.
From the first equation, we conclude that
y
=
x
−
4
x
. Substitution into the second equation
yields
(
x
−
4
x
) =
−
2
x
+ (
x
−
4
x
)
⇒
x
−
5
x
+ 6
x
= 0
.
The characteristic equation,
r
2
−
5
r
+ 6 = 0, has roots
r
= 2
,
3, and so a general solution is
x
(
t
) =
c
1
e
2
t
+
c
2
e
3
t
⇒
y
(
t
) =
(
c
1
e
2
t
+
c
2
e
3
t
)
−
4
(
c
1
e
2
t
+
c
2
e
3
t
)
=
−
2
c
1
e
2
t
−
c
2
e
3
t
.
We find constants
c
1
and
c
2
from the initial condition.
1 =
x
(0) =
c
1
e
2(0)
+
c
2
e
3(0)
=
c
1
+
c
2
,
0 =
y
(0) =
−
2
c
1
e
2(0)
−
c
2
e
3(0)
=
−
2
c
1
−
c
2
⇒
c
1
=
−
1
,
c
2
= 2
.
Therefore, the answer to this problem is
x
(
t
) =
−
e
2
t
+ 2
e
3
t
,
y
(
t
) = 2
e
2
t
−
2
e
3
t
.
21.
To apply the elimination method, we write the system using operator notation:
D
2
[
x
]
−
y
= 0
,
−
x
+
D
2
[
y
] = 0
.
(5.5)
Eliminating
y
by applying
D
2
to the first equation and adding to the second equation gives
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Characteristic polynomial, ﬁrst equation

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