Exercise Set 5
Math 4027
Due:
May 5, 2005
1. Solve the following differential equations:
(a)
y
0
=
x
2
/y
.
I
Solution.
The equation is separable, so rewrite it in the form
yy
0
=
x
2
or in
differential form
y dy
=
x
2
dx
and integrate to get an implicit equation
y
2
2
=
x
3
3
+
C.
A solution is any function
ϕ
(
x
) implicitly defined by this equation on some interval
I
, so long as
ϕ
(
x
)
6
= 0 for
x
∈
I
.
J
(b)
y
0
+
y
2
sin
x
= 0
I
Solution.
Separate the variables to get
y

2
y
0
=

sin
x
or in differential form
y

2
dy
=

sin
x dx
and integrate to get an implicit equation

1
y
= cos
x
+
C.
This can be solved explicitly for
y
on any interval
I
for which cos
x
+
C
6
= 0 for
all
x
∈
I
to get
y
=

1
cos
x
+
C
.
There is also one constant solution, namely
y
= 0, which is lost in the process of
separating the variables.
J
2. Solve the initial value problem
y
0
=
3
x
2
3
y
2

4
,
y
(1) = 0
,
and determine the interval in which the solution is valid. (
Hint:
To find the interval
of definition, look for points where the integral curve has a vertical tangent.)
I
Solution.
Separate the variables to get (3
y
2

4)
y
0
= 3
x
2
or in differential form
(3
y
2

4)
dy
= 3
x
2
dx
and integrate to get an implicit equation
y
3

4
y
=
x
3
+
C,
where
C
is an integration constant. The initial value
y
(1) = 0 means that
y
= 0 when
x
= 1, so this implies that
C
=

1 and we get that the initial value problem is defined
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Adkins
 Differential Equations, Equations, Constant of integration, Coddington

Click to edit the document details