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Exercises 5.2
and the general solution,
x
(
t
), is
x
(
t
)=
x
h
(
t
)+
x
p
(
t
C
1
e
r
1
t
+
C
2
e
r
2
t
+20
.
From the frst equation in (5.18) we fnd
y
(
t
)
=
100
·{
(
D
+0
.
07)[
x
]
−
1
.
2
}
= 100
dx
dt
+7
x
(
t
)
−
120
=
100
±
r
1
C
1
e
r
1
t
+
r
2
C
2
e
r
2
t
²
±
C
1
e
r
1
t
+
C
2
e
r
2
t
²
−
120
=
³
2
−
√
7
´
C
1
e
r
1
t
+
³
2+
√
7
´
C
2
e
r
2
t
.
The initial conditions imply
0=
x
(0) =
C
1
+
C
2
,
20 =
y
(0) =
(
2
−
√
7
)
C
1
+
(
√
7
)
C
2
⇒
C
1
+
C
2
=
−
20
,
(
2
−
√
7
)
C
1
+
(
√
7
)
C
2
=0
⇒
C
1
=
−
µ
10 +
20
√
7
¶
,C
2
=
−
µ
10
−
20
√
7
¶
.
Thus the solution to the problem is
x
(
t
−
µ
10 +
20
√
7
¶
e
r
1
t
−
µ
10
−
20
√
7
¶
e
r
2
t
+20(kg)
,
y
(
t
30
√
7
e
r
1
t
−
30
√
7
e
r
2
t
.
33.
Since no solution ﬂows in or out o± the system ±rom the tank B, we conclude that the solution
ﬂows ±rom the tank B to the tank A with the same rate as it does ±rom A to B, that is, 1 L/min.
Furthermore, the solution ﬂows in and out o± the tank A with the same rate, 4 L/min, and so
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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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