Math 20D
Practice Exam I Solutions
1. (a) (10 points) Find the general solution to the differential equation
t
3
y
0
+ 4
t
2
y
=
e

t
.
Rewrite as
y
0
+
4
t
y
=
e

t
t
3
. This is a linear firstorder equation, so next
compute the integrating factor
μ
=
e
R
(4
/t
)
dt
=
e
4 ln
t
=
t
4
.
Multiply
through by
t
4
to get
t
4
y
0
+ 4
t
3
y
=
te

t
. The LHS is
d
dt
(
t
4
y
), so taking
the antiderivative of both sides gives
t
4
y
=

te

t

e

t
+
C
. So the
general solution is
y
(
t
) =

e

t
t
3

e

t
t
4
+
C
t
4
.
(b) (8 points) Find the particular solution to the differential equation
in part (a) with initial value
y
(1) = 0.
y
(1) =

e

1

e

1
+
C
= 0, so
C
= 2
e

1
.
The solution is
y
(
t
) =

e

t
t
3

e

t
t
4
+
2
e

1
t
4
.
(c) (2 points) What is the largest interval on which the solution you
found in part (b) is defined?
The solution is defined on the interval (0
,
∞
).
2. Consider the autonomous differential equation
dy
dt
=
f
(
y
)
,
where
f
(
y
) = (9

y
2
)(4

y
2
).
(a) (10 points) Draw a phase line, and label each equilibrium solution
as stable, unstable, or semistable.
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 Spring '06
 Mohanty
 Math, Derivative

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