Math 2220
Problem Set 9 Solutions
Spring 2010
Section 3.3
:
1.
For the vector field
F
(
x,y
) = (
−
x,y
) do the following things:
(a) Sketch the vector field.
Solution.
Along the
x
axis we have
F
(
x,
0) = (
−
x,
0),
pointing toward the origin. Along the
y
axis we have
F
(0
,y
) = (0
,y
), pointing away from the origin. This
gives a start in sketching the vector field, as shown in
the figure. Looking on the lines
y
=
x
and
y
=
−
x
we
get
F
(
x,x
) = (
−
x,x
) and
F
(
x,
−
x
) = (
−
x,x
), which
gives us a bit more, enough to predict what the rest
will look like.
(b) Determine the flow lines by solving the differential equations
dx
dt
=
−
x
and
dy
dt
=
y
.
Your answer should be functions
x
(
t
) and
y
(
t
) each of which depends on an arbitrary
constant.
Solution.
By inspection,
dx
dt
=
−
x
has solutions
x
=
Ae

t
for
A
an arbitrary constant.
More systematically, we can rewrite the equation
dx
dt
=
−
x
as
dx
x
=
−
dt
. Integrating both
sides of this equation gives ln
x
=
−
t
+
C
, which implies
x
=
e
C
e

t
=
Ae

t
. In a similar
way the equation
dy
dt
=
y
gives
y
=
Be
t
for
B
an arbitrary constant.
(c) Check that the flow lines are hyperbolas by eliminating the variable
t
from your
formulas in part (b).
Solution.
We have
xy
=
Ae

t
Be
t
=
AB
. This can be written as
xy
=
C
for
C
=
AB
.
These curves are the hyperbolas
y
=
C/x
.
(d) Find the values of the arbitrary constants that give the flow line
c
(
t
) such that
c
(0) =
(2
,
1).
Solution.
For
t
= 0 the equation (
x,y
) = (
Ae

t
,Be
t
) = (2
,
1) gives (
A,B
) = (2
,
1) so
A
= 2 and
B
= 1 and we have
c
(
t
) = (2
e

t
,e
t
).
2.
Sketch the vector field
F
(
x,y
) = (
x,x
2
), determine the flow lines, and show the flow
lines are parabolas.
Solution.
Notice that the formula (
x,x
2
) is indepen
dent of
y
, so the vector field looks the same on each
vertical line. Thus is suffices to sketch the vector field
along the
x
axis and then translate the result verti
cally up and down.
This leads to the figure at the
right.