HOMEWORK 1: MATH 265, L Keshet
Due in class on September 22, 2010
Problem 1:
In each case, solve for
y
(
t
):
(a)
y
′
+ 3
y
= 2
e
t/
2
with
y
(0) = 1.
(b)
y
′

4
y
=
t
with
y
(1) = 0.
(c)
ty
′
+ 2
y
= sin
t
with
y
(
π/
2) = 0.
(d)
t
2
y
′
+ 2
ty
=
t
3
+ 1 with
y
(1) = 1.
Solution to Problem 1:
(a) The integrating factor is
μ
(
t
) =
e
3
t
. The solution is
y
(
t
) =
4
7
e
7
t/
2
+
Ce
−
3
t
. Using the initial condition,
the constant is found and the solution is
y
(
t
) =
4
7
e
7
t/
2
+
3
7
e
−
3
t
.
(b) The integrating factor is
μ
(
t
) =
e
−
4
t
.
Integration by parts is needed to find that the solution is
y
(
t
) =

1
16

1
4
+
Ce
4
t
Using the initial condition, we find the constant. The result is
y
(
t
) =

1
16

1
4
+
5
16
e
4
e
4
t
(c) The integrating factor is
μ
(
t
) =
t
2
(
typo corrected
). Integration by parts is needed to find that the
solution is
y
(
t
) =
sin(
t
)
t
2

cos(
t
)
t
+
C
t
2
The constant is found to be
C
=

1 (
typo corrected
).
(d) The LHS is actually simply
d
(
t
2
y
(
t
))
/dt
so we do not need an integrating factor.
We can simply
integrate both sides. We get
y
(
t
) =
1
t
+
t
2
4
+
C
1
t
2
.
Using the initial condition, we find that the constant
is
C
=

1
/
4.
Problem 2:
Draw the direction fields for each of the below and sketch the solution curve corresponding to
the indicated initial condition.
(a)
y
′
=
y

2 cos(
t
) + 1,
y
(0) = 0
(b)
y
′
=
√
y

ty
,
y
(0) = 2
Solution to Problem 2:
For part (a) we note that the directions of tangent lines to solution curves will
all be flat (have slope zero) whenever 0 =
y

2 cos(
t
) + 1.
We can sketch this set of points.
It may be
easiest to first rewrite it as
y
= 2 cos(
t
)

1, as we see that it is a cosine of amplitude 2 shifted down
the y axis by 1 (dotted line in Fig 1(a)). All along this locus, we sketch horizontal “dashes”. This locus
separates portion of the yt plane for which solutions are increasing (i.e. where
y
′
(
t
)
>
0) from the portion
1
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where solutions are decreasing (
y
′
(
t
)
<
0).
We can go further by noting that the slope of tangent lines
will have a value
M
all along the locus
M
=
y

2 cos(
t
) + 1. We can similarly sketch such sets of points
(
y
= 2 cos(
t
) +
M

1) and assign to them directions of slope
M
for various values of
M
to get a fairly good
picture.
The (computer generated) direction field, together with one solution curve satisfying the initial
condition is shown in Fig 1(a).
For part (b) the equation can be written as
y
′
=
√
y

ty
=
√
y
(1

t
√
y
). First we see that
y
always
has to be positive for real values, so only the first two quadrants are relevant. If
t <
0 then (1

t
√
y
) is
positive. This means arrows have positive slope in the 2nd quadrant. On the curve (1

t
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 Winter '10
 LEAHKESHET
 Math, Differential Equations, Equations, Derivative, Constant of integration, initial condition

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