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Chapter 5
(b)
r
=
−
1
±
p
1
2
−
4(
−
6)
2
=
−
1
±
5
2
=
−
3
,
2
⇒
D
2
+
D
−
6=(
D
+3)(
D
−
2).
(c)
r
=
−
9
±
p
9
2
−
4(
−
5)2
4
=
−
9
±
11
4
=
−
5
,
1
/
2
⇒
2
D
2
+9
D
−
5=(
D
+ 5)(2
D
−
1).
(d)
r
=
±
√
2
⇒
D
2
−
2=(
D
+
√
2)(
D
−
√
2).
EXERCISES 5.3: Solving Systems and Higher–Order Equations Numerically, page 261
1.
We isolate
y
0
(
t
) frst and obtain an equivalent equation
y
0
(
t
)=3
y
(
t
)
−
ty
0
(
t
)+
t
2
.
Denoting
x
1
:=
y
,
x
2
:=
y
0
we conclude that
x
0
1
=
y
0
=
x
2
,
x
0
2
=(
y
0
)
0
=
y
0
=3
y
−
ty
0
+
t
2
x
1
−
tx
2
+
t
2
,
with initial conditions
x
1
(0) =
y
(0) = 3,
x
2
(0) =
y
0
(0) =
−
6. ThereFore, given initial value
problem is equivalent to
x
0
1
=
x
2
,
x
0
2
x
1
−
tx
2
+
t
2
,
x
1
(0) = 3
,x
2
(0) =
−
6
.
3.
Isolating
y
(4)
(
t
), we get
y
(4)
(
t
)=
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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