ACM 95b/100b
Solutions for Problem Set 1
Sean Mauch
[email protected]
January 18, 2002
1
Problem 1
Problem.
Find the general solution and plot some integral curves for each of the following ODEs.
What is the behavior of the solution as
t
→ ∞
?
1.
ty
0

y
=
t
2
2.
y
0
+
y
cos
t
= sin
t
cos
t
Solution.
1. We consider the differential equation
ty
0

y
=
t
2
.
We can solve this differential equation by inspection.
We guess a particular solution of the
form
y
p
=
at
2
.
By substituting into the differential equation we see that
y
p
=
t
2
.
We see
that
y
h
=
t
is a homogeneous solution. Now we have the general solution of the differential
equation.
y
=
t
2
+
ct
We could also solve this problem by finding an integrating factor.
y
0

y
t
=
t
I
=
exp

Z
1
t
d
t
= e

ln

t

=
1

t

We choose
I
= 1
/t
as the integrating factor.
d
d
t
y
t
= 1
y
t
=
t
+
c
y
=
t
2
+
ct
The solution, plotted in Figure 1, is unbounded as
t
→ ∞
.
2. We consider the differential equation
y
0
+
y
cos
t
= sin
t
cos
t.
1
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2
1
1
2
2
1.5
1
0.5
0.5
1
1.5
2
Figure 1: The integral curves of
ty
0

y
=
t
2
.
Again we can solve this differential equation by inspection.
We see that
y
p
= sin
t

1 is a
particular solution and
y
h
= e

sin
t
is a homogeneous solution. The general solution of the
differential equation is
y
= sin
t

1 +
c
e

sin
t
.
We could also solve this problem by finding an integrating factor.
I
= exp
Z
cos
t
d
t
= e
sin
t
d
d
t
(
e
sin
t
y
)
= e
sin
t
sin
t
cos
t
e
sin
t
y
=
Z
e
sin
t
sin
t
cos
t
d
t
+
c
e
sin
t
y
=
Z
sin
t
e
x
x
d
x
+
c
e
sin
t
y
= [(
x

1) e
x
]
sin
t
+
c
y
= sin
t

1 +
c
e

sin
t
The solution, plotted in Figure 2, is bounded but does not have a limit as
t
→ ∞
.
2
Problem 2
Problem.
Solve the following linear initial value problems and in each case describe the interval
on which the solution is defined.
1.
y
0
+ 2
xy
= e

x
2
,
y
(0) =

1
2.
ty
0
+ 2
y
=
t
2

t
+ 1,
y
(1) = 1
/
2
Solution.
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 Winter '09
 NilesA.Pierce
 Constant of integration, [email protected]tech.edu

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