AM 3815  Solutions to Assignment 1
1.
Exercise 2, page 9, Strauss
Following the hint, we substitute
v
=
u
y
into the equation 3
u
y
+
u
xy
= 0 to obtain
3
v
+
v
x
= 0
⇔
v
x
=

3
v
.
It follows that
v
=
f
(
y
) e

3
x
=
u
y
, and antidifferentiating this equation we find
u
=
Z
f
(
y
)
dy
+
g
(
x
)
e

3
x
,
and so the solution is
u
(
x
,
y
) =
F
(
y
)e

3
x
+
G
(
x
).
2.
Exercise 5, page 10, Strauss
We are asked to solve
xu
x
+
yu
y
= 0.
To apply the method of characteristics we first solve the initial value problem,
dx
dt
=
x
with
x
(0) =
x
0
dy
dt
=
y
with
y
(0) =
y
0
to obtain
x
(
t
) =
x
0
e
t
,
y
(
t
) =
y
0
e
t
.
Now we define
U
(
t
) =
u
(
x
(
t
),
y
(
t
)). Differentiating with respect to
t
we find
U
0
(
t
) =
x u
x
+
y u
y
= 0
and so
U
(
t
) = something that is constant with respect to
t
.
When we translate back to
x
,
y
coordinates, then, we have
u
(
x
,
y
) =
f
, but
f
cannot depend
on
x
and
y
in a way that would dependent on
t
. From above we see that
y
(
t
)
/
x
(
t
) does not
depend on
t
and so we can assert
u
(
x
,
y
) =
f
(
y
/
x
).
3.
Exercise 6, page 10, Strauss
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
We can again apply the Method of Characteristics. We first solve the initial value problem
dx
dt
=
√
1

x
2
with
x
(0) =
x
0
dy
dt
=
1
with
y
(0) =
y
0
to get
x
(
t
) = sin(
t
+
x
0
),
y
(
t
) =
t
+
y
0
and so if we combine
x
and
y
as
(
y

y
0
)

(arcsin
x

x
0
) = (
y

arcsin
x
) + (
x
0

y
0
)
we get a composite variable, namely (
y

arcsin
x
), that is not dependent on
t
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 llll
 Constant of integration, Strauss, dt dy dt, dx dt dy

Click to edit the document details