APPM 2360
Final Exam: May 2, 2005
1. (25 points)
(a) Find the general solution of
y
′′
= 2
y
′
+ 3
y
(b) Convert the equation in (a) to a firstorder system of equations and write the system in
matrixvector form
(c) Find the eigenvalues of the matrix in (b). How do they relate to your answer to (a)?
(d) Find the general solution to the system in (b). How does this solution — in particular,
the individual components of the vector solution — relate to your answer to (a)?
Solutions:
(a)
y
′′

2
y
′

3
y
= 0 so try
y
=
e
rt
to get char. polynomial:
r
2

2
r

3 = (
r

3)(
r
+ 1) =
0
⇒
r
= 3
,

1. Hence general solution is
y
(
t
) =
c
1
e
3
t
+
c
2
e
−
t
.
(b) Define
v
=
y
′
, so
y
′
=
v
and
v
′
=
y
′′
= 2
y
′
+ 3
y
= 3
y
+ 2
v
. The system is
y
′
=
v
v
′
=
3
y
+ 2
v
⇔
parenleftBigg
y
v
parenrightBigg
′
=
parenleftBigg
0
1
3
2
parenrightBiggparenleftBigg
y
v
parenrightBigg
(c) det(
A

λI
) =
λ
2

(0 + 2)
λ
+ (0

3) =
λ
2

2
λ

3 = (
λ

3)(
λ
+ 1) = 0
⇒
λ
= 3
,

1,
which are the same are the roots of the char. poly. in (a).
(d) Finding the eigenvectors:
λ
= 3
⇒
(
A

λI
) =
parenleftBigg

3
1
3

1
parenrightBigg
⇒
v
=
parenleftBigg
1
3
parenrightBigg
;
λ
=

1
⇒
(
A

λI
) =
parenleftBigg
1
1
3
3
parenrightBigg
⇒
v
=
parenleftBigg
1

1
parenrightBigg
. Hence the general solution is
parenleftBigg
y
v
parenrightBigg
=
c
1
e
3
t
parenleftBigg
1
3
parenrightBigg
+
c
1
e
−
t
parenleftBigg
1

1
parenrightBigg
The first component of the vector solution is
y
(
t
) =
c
1
e
3
t
+
c
2
e
−
t
, which is the same as
found in (a). The second component is
v
(
t
) = 3
c
1
e
3
t

c
2
e
−
t
, which is the derivative of
y
(
t
) (as it should be, since
v
was defined as
y
′
).
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2. (20 points) Find the general solution of
y
′′

4
y
′
+ 4
y
=
f
(
x
), where
(a)
f
(
x
) = 3
e
−
x
(b)
f
(
x
) = 2 sin(2
x
)
Solutions:
(a) Taking the homogeneous part:
y
′′

4
y
+ 4
y
=, try
y
=
e
rx
and get char. polynomial:
r
2

4
r
+ 4 = (
r

2)
2
= 0
⇒
r
= 2 (repeated).
Hence
y
h
(
t
) =
c
1
e
2
x
+
c
2
te
2
x
.
Use
Undetermined Coefficients to find a particular solution:
y
p
=
Ae
−
x
⇒
y
′
p
=

Ae
−
x
and
y
′′
p
=
Ae
−
x
.
Hence
y
′′
p

4
y
′
p
+ 4
y
p
=
e
−
x
(
A
+ 4
A
+ 4
A
) = 9
Ae
−
x
= 3
e
−
x
⇒
A
=
1
3
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '07
 WILLIAMHEUETT
 Linear Algebra, Trigraph, Elementary algebra, general solution, real solutions

Click to edit the document details