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MATH 141B
QUIZ#7 Solutions
Date: Tues March 18, 2008
1.
V
= 100 and
F
= 25
,
so
F/V
= 1
/
4
.
Since
I
=
c
·
F
= (0
.
1)(25) = 2
.
5
,
the diferential
equation is
y
′

1
4
y
+ 2
.
5. In standard Form, we get
y
′
+
1
4
y
=
5
2
.
Now,
F
(
t
) =
i
1
4
dt
=
1
4
t,
so the integrating Factor is
e
t/
4
.
Multiplying by
G
gives,
e
t/
4
y
′
+
1
4
e
t/
4
y
=
5
2
e
t/
4
.
By the
product rule For diferentiation, this becomes (
ye
t/
4
)
′
=
5
2
e
t/
4
.
Integrating, and we get
ye
t/
4
=
10
e
t/
4
+
C.
We don’t need to ±nd
C
since it will vanish in the limit. Multiplying by
e
−
t/
4
gives
y
(
t
) = 10 +
Ce
−
(
t/
4)
.
²inally,
y
∞
= lim
t
→∞
y
(
t
) = 10
.
ThereFore, A) is the answer.
2.
In standard Form, the diferential equation is
s
′
+
(
ln 2
30
)
s
=
1
2
. Now,
F
(
t
) =
i
ln 2
30
dt
=
ln 2
30
t.
ThereFore, the integrating Factor is
G
(
t
) =
e
(ln 2
/
30)
t
.
Multiplying the D.E. by
G
(
t
) gives
[
s
·
e
(ln 2
/
30)
t
]
′
=
1
2
e
(ln 2
/
30)
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This note was uploaded on 03/30/2008 for the course MATH 141B taught by Professor Marcfabbri during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 MARCFABBRI
 Math, Logic

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