Name:
Homework 3 Solutions
Math 285
Write the general solution as
y
=
A
+
Be

t
+
Ce
t
and apply the initial conditions. We get the system
y
(0) =
A
+
B
+
C
= 2
y
0
(0) =

B
+
C
= 3
y
00
(0) =
B
+
C
= 4
and we find that
C
= 7
/
2
, B
= 1
/
2
, A
=

2 so
y
=

2 + 1
/
2
e

t
+ 7
/
2
e
t
Problem 4
[2pts]
Compute the Wronskian of the functions
y
1
(
t
) = 1
y
2
(
t
) =
e

t
y
3
(
t
) =
e
t
Answer:
The Wronskian is given by
W
(
y
1
, y
2
, y
3
) =

2
Problem 5
[2pts]
Given what you know from problem 3 what does Abel's theorem
tell you about the Wronskian of
y
1
(
t
) = 1
y
2
(
t
) =
e

t
y
3
(
t
) =
e
t
You should not need to calculate any determinants.
Answer:
Since
y
1
, y
2
, y
3
are all solutions to
d
3
y
dt
3

dy
dt
= 0
it follows that the Wronskian
W
satisfies the equation
dW
dt
= 0
so we know that the Wronskian is constant. The constant itself can only be found by taking the determinant.
2