The Wronskian satisfies:
W
[
y
1
, y
2
](
t
) =
cos(2
t
)
sin(2
t
)

2 sin(2
t
)
2 cos(2
t
)
= 2(cos
2
(2
t
) + sin
2
(2
t
)) = 2
.
The variation of parameters gives
y
p
(
t
)
=

cos(2
t
)
Z
t
8 sin(2
s
) csc(2
s
)
2
ds
+ sin(2
t
)
Z
t
8 cos(2
s
) csc(2
s
)
2
ds
=

4 cos(2
t
)
Z
t
1
·
ds
+ 4 sin(2
t
)
Z
t
cos(2
s
)
sin(2
s
)
ds
=

4
t
cos(2
t
) + 2 sin(2
t
) ln

sin(2
t
)

The general solution satisfies:
y
(
t
) =
c
1
cos(2
t
) +
c
2
sin(2
t
)

4
t
cos(2
t
) + 2 sin(2
t
) ln

sin(2
t
)

.
e. The differential equation,
y
00
+ 4
y
0
+ 4
y
=
e

2
t
t
2
,
requires the variation of parameters. The characteristic equation is
λ
2
+ 4
λ
+ 4 = 0, so
λ
=

2
(repeated root). Thus, the homogeneous solution is
y
c
(
t
) = (
c
1
+
c
2
t
)
e

2
t
.
Subscribe to view the full document.
The Wronskian satisfies:
W
[
y
1
, y
2
](
t
) =
e

2
t
te

2
t

2
e

2
t
(1

2
t
)
e

2
t
=
e

4
t
.
The variation of parameters gives
y
p
(
t
)
=

e

2
t
Z
t
se

2
s
e

2
s
s
2
e

4
s
ds
+
te

2
t
Z
t
e

2
s
e

2
s
s
2
e

4
s
ds
=

e

2
t
Z
t
1
s
ds
+
te

2
t
Z
t
1
s
2
ds
=

e

2
t
ln

t
 
e

2
t
The general solution satisfies:
y
(
t
) = (
c
1
+
c
2
t
)
e

2
t

e

2
t
(ln

t

+ 1)
.
f. The differential equation,
t
2
y
00
+ 7
ty
0
+ 5
y
= 6
t
, requires the variation of parameters.
The
left hand side is a CauchyEuler equation, so attempt solutions of the form
y
(
t
) =
t
λ
, leaving a
characteristic equation
λ
(
λ

1) + 7
λ
+ 5 =
λ
2
+ 6
λ
+ 5 = 0 or
λ
1
=

5 and
λ
2
=

1. Thus, the
homogeneous solution is
y
c
(
t
) =
c
1
t

5
+
c
2
t

1
.
The Wronskian satisfies:
W
[
y
1
, y
2
](
t
) =
t

5
t

1

5
t

6

t

2
= 4
t

7
.
Note that the nonhomogeneous term is written
g
(
t
) =
6
t
. The variation of parameters gives
y
p
(
t
)
=

t

5
Z
t
6(
s

1
)
4
s
(
s

7
)
ds
+
t

1
Z
t
6(
s

5
)
4
s
(
s

7
)
ds
=

3
2
t

5
Z
t
s
5
ds
+
3
2
t

1
Z
t
sds
=

3
2
t
5
t
6
6
+
3
2
t
t
2
2
=
1
2
t
The general solution satisfies:
y
(
t
) =
c
1
t

5
+
c
2
t

1
+
1
2
t.
2. The differential equation is:
(1

t
)
y
00
+
ty
0

y
= 2(
t

1)
2
e

t
or
y
00
+
t
1

t
y
0

1
1

t
y
=

2(
t

1)
e

t
.
Since
y
1
(
t
) =
t
, then
y
0
1
(
t
) = 1 and
y
00
1
(
t
) = 0. Substituting into the RHS of the equation above
gives:
0 +
t
1

t

1
1

t
t
= 0
,
so satisfies the homogeneous equation.
Similarly, if
y
2
(
t
) =
e
t
, then
y
0
2
(
t
) =
e
t
and
y
00
2
(
t
) =
e
t
.
Substituting into the RHS of the equation above gives:
e
t
1

t
1

t
+
te
t
1

t

e
t
1

t
= 0
,
so
y
2
(
t
) satisfies the homogeneous problem. Solving the general problem requires finding a particular
solution using the variation of parameters. The Wronskian satisfies:
W
[
y
1
, y
2
](
t
) =
t
e
t
1
e
t
= (
t

1)
e
t
.
The variation of parameters gives
y
p
(
t
)
=

t
Z
t

e
s
(2(
s

1)
e

s
)
(
s

1)
e
s
ds
+
e
t
Z
t

s
(2(
s

1)
e

s
)
(
s

1)
e
s
ds
=
t
Z
t
2
e

s
ds

e
t
Z
t
2
se

2
s
ds
=

2
te

t
+
t
+
1
2
e

t
=
1
2

t
e

t
The general solution satisfies:
y
(
t
) =
c
1
t
+
c
2
e
t
+
1
2

t
e

t
.
Subscribe to view the full document.
 Fall '08
 staff