Math 293 Solutions to Problem Set 7.
§
2.3 #3
. The equation
x
dx
dt
+
t
2
x
= sin
t
is nonlinear because of the product
x
dx
dt
. Rewrit
ing it as
dx
dt
=
sin
t
x

t
2
we see that it is not separable because the righthand side cannot
be factored as a product
f
(
x
)
g
(
t
).
#4
. The equation is linear since it can be written
e
t dy
dt
+ (ln
t
)
y
= 3
t
. It is not separable
since in the form
dy
dt
=
3
t

y
ln
t
e
t
the right side cannot be factored as
f
(
y
)
g
(
t
).
#15
.
Rewrite the equation as
y
0
+
x
x
2
+1
y
=
x
x
2
+1
.
To compute an integrating factor
we compute
R
x
x
2
+1
dx
= ln
√
x
2
+ 1, so an integrating factor is
√
x
2
+ 1.
The differen
tial equation then takes the form (
y
√
x
2
+ 1)
0
=
x
x
2
+1
.
Integrating both sides, we get
y
√
x
2
+ 1 =
R
x
√
x
2
+1
dx
=
√
x
2
+ 1 +
C
. Solving this for
y
gives
y
= 1 +
C
√
x
2
+1
.
#20
. Write the equation as
y
0
+
3
x
y
= 3
x

2. An integrating factor is
e
R
3
x

1
dx
=
e
3 ln
x
=
x
3
. Multiplying both sides of the equation by
x
3
then gives (
x
3
y
)
0
= 3
x
4

2
x
3
. Integrate
both sides to get
x
3
y
=
3
x
5
5

x
4
2
+
C
.
The initial condition
y
(1) = 1 yields the value
C
=
9
10
. Solving for
y
, we get
y
=
3
x
5
5

x
2
+
9
10
x

3
.
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 '07
 TERRELL,R
 Math, Differential Equations, Equations, Quadratic equation, Elementary algebra, Constant of integration, Boundary value problem

Click to edit the document details