MATH 3103
Fall 2011
Differential Equations
Homework Set 5
Some Solutions
The following solutions are not necessarily complete, but should highlight the main points for most
questions. If you have any further questions regarding the problems and solutions, feel free to discuss
them on WebCT.
Solutions marked by
*
are taken from the publisher’s solutions for the textbook by Boyce and DiPrima,
9th edition (Wiley).
Section 2.4:
*
Section 2.4 #
6
:
[Recall that cot
t
= 1
/
tan
t
= cos
t/
sin
t
, so the discontinuities of cot
t
occur where sin
t
= 0, that
is, for
t
=
kπ
, where
k
∈
Z
(that is,
k
is an integer). Thus the possible intervals of existence are
(0
,
1), (1
, π
), and (
kπ,
(
k
+1)
π
) for
k
6
= 0. To find the interval of existence for the given initial value
problem, we need to determine which of these intervals contains the initial condition at
t
= 2; the
answer is (1
, π
).]
*
Section 2.4 #
12
:
[That is, the hypotheses of the existenceuniqueness theorem are satisfied everywhere
except
for
t
=
kπ
,
k
∈
Z
and
y
=

1.]
*
Section 2.4 #
15
:
[For
y
0
6
= 0, the interval of existence is (

1
2
y

2
0
,
∞
).]
•
Section 2.4 #
21
:
As derived in class (see also Section 2.4 Example 3), the initial value problem
y
0
=
y
1
/
3
,
y
(0) = 0
has infinitely many solutions, namely
y
(
t
)
≡
0 and, for each
t
0
≥
0,
y
(
t
) =
(
0
,
0
≤
t < t
0
,
±
2
3
(
t

t
0
)
3
/
2
,
t
≥
t
0
.
1. To get
y
(1) = 1, we need a positive solution (use “+”) with 1 =
2
3
(1

t
0
)
3
/
2
or
2
3
t
0
=
2
3

1,
which is not possible for
t
0
≥
0. Hence there is
no
solution passing through the point (1
,
1).
2. To get
y
(2) = 1, we need 1 =
2
3
(2

t
0
)
3
/
2
or
2
3
t
0
=
4
3

1 =
1
3
, so
t
0
= 1
/
2.
3. Considering all (positive and negative) solutions, we have
y
(2) =
±
2
3
(2

t
0
)
3
/
2
for 0
≤
t
0
≤
2 and
y
(2) = 0 for
t
0
>
2. The largest possible values (in absolute value) occur for
t
0
= 0, for
which

y
(2)

= (4
/
3)
3
/
2
. Hence the set of possible values is
y
(2)
∈
[

(4
/
3)
3
/
2
,
(4
/
3)
3
/
2
].