Nonbook problems
1. Verify that for any nonzero constant
b,
the function
f
(
x
) =
1
b
cosh(
bx
) satisfies the differential
equation
d
2
y
dx
2

b
v
u
u
t
1 +
dy
dx
!
2
= 0
.
(Recall that the function “cosh” is defined by cosh(
x
) =
1
2
(
e
x
+
e

x
).)
2. Verify that on the interval

2
< x <
2, the two continuous functions
y
(
x
) =
√
4

x
2
and
y
(
x
) =

√
4

x
2
, obtainable from the implicit solution
x
2
+
y
2
= 4 of the differential equation
x
+
y dy/dx
= 0, are (explicit) solutions of this differential equation.
3. Find the general solution of
dy
dx
=
x
sin
x
ln
y
.
4. Find the general solution of
dy
dx
=
tan

1
x
ye
2
y
.
(
Notational reminder
: “tan

1
” denotes the inversetangent function, also called arctangent, and
also written “arctan”. It does
not
denote the reciprocal of the tangent function, which is the
cotangent function “cot”.)
5. Consider the initialvalue problem
dx
dt
+
x
2
=
x,
x
(0) =
x
0
(1)
(This is the same DE as in exercise 13 of Section 2.2, which you should do prior to starting this
exercise.)
For each of the following values of
x
0
, find both the solution of the IVP and the domain of
the solution. The answers to “What is the domain of the solution?” are given below, but
they
are not obvious
. Do
not
expect to be able to find a quick way for guessing what the answers will
be based on
x
0
. To get these answers you will have to
solve the IVP first
—you should get an
explicit formula for
x
(
t
)—and then, from your formula, figure out the domain of the solution
(remembering that only
intervals
are allowed as solutions of ODEs, even if the formula you
write down has a larger domain).
(a)
x
0
=
1
2
Answer for domain:
(
∞
,
∞
)
(i.e.
∞
< t <
∞
)
(b)
x
0
= 2
Answer for domain:
(

ln 2
,
∞
)
(i.e.

ln 2
< t <
∞
)
(c)
x
0
=

2
Answer for domain:
(
∞
,
ln(
3
2
)
(i.e.
∞
< t <
ln(
3
2
)
)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(d)
x
0
= 0
Answer for domain:
(
∞
,
∞
)
(e)
x
0
= 1
Answer for domain:
(
∞
,
∞
)
6. Let
p
be a function that is differentiable on the whole real line, and consider the separable
differential equation
dy
dx
=
p
(
y
)
.
(2)
(Here, the function
g
(
x
) that you’re used to seeing is just the constant function 1.) Note
that if we rename the variables
x, t
in the previous exercise to
y, x
respectively, the ODE in
that exercise can be rewritten in this form, with
p
(
y
) =
y

y
2
.
(a) Show that the family of all solutions of (2) is
translationinvariant
in the following sense:
if
y
=
φ
(
x
) is a solution on an interval
a < x < b
, and
k
is any constant, then
y
=
φ
(
x

k
) is
a solution on the interval
a
+
k < x < b
+
k
. (Said another way: horizontally translating the
graph of a solution by any amount, you get the graph of another solution.)
(b) Using the Fundamental Existence/Uniqueness Theorem for FirstOrder InitialValue
Problems (Theorem 1 on p. 12 in the textbook), show that for every point (
x
0
, y
0
)
∈
R
2
, the
initialvalue problem
dy
dx
=
p
(
y
)
,
y
(
x
0
) =
y
0
,
(3)
has a unique solution on some open interval containing
x
0
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 TUNCER
 Derivative, Continuous function, dy, Fupper, open quadrant

Click to edit the document details