B
˙
ILKENT UNIVERSITY
Department of Mathematics
MATH 225,
LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS,
Solution of Homework set
1
# 17
U. Mu˘gan
July 16, 2008
1)
a)
Substituting
y
=
e
rx
into the D.E., we get the C.E.
r
2

2
r
+ 1 = 0 which gives
r
1
=
r
2
= 1.
Then two L.I. solutions
y
1
=
e
x
and
y
2
=
xe
x
. So the general solution is
y
(
x
) =
c
1
e
x
+
c
2
xe
x
.
Or, we can use the method of reduction of order if
y
1
=
e
x
, then the second L.I. solution is of the
form
y
2
=
y
1
v
(
x
) =
e
x
v
(
x
). D.E yields
v
00
= 0. Solving the D.E. for
v
gives
v
(
x
) =
c
1
+
c
2
x,
c
1
, c
2
=
constant. The general solution of the given D.E. can be obtained from
y
=
e
x
v
(
x
) .
b)
Substituting
y
=
e
rx
into the D.E. we get the C.E. 25
r
2

20
r
+ 4 = 0 which may be written as
(5
r

2)
2
= 0. The roots of the C.E. are
r
1
=
r
2
= 2
/
5.
Thus,
y
(
x
) =
c
1
e
2
x/
5
+
c
2
xe
2
x/
5
,
c
1
, c
2
= constant.
c)
Substituting
y
=
e
rx
into the D.E. we get the C.E.
r
2

2
r
+ 10 = 0 and hence the roots are
r
1
= 1 + 3
i,
r
2
= 1

3
i
. Thus,
y
(
x
) =
c
1
e
x
cos 3
x
+
c
2
e
x
sin 3
x,
c
1
, c
2
= constant.
2)
a)
The C.E. is
r
2
+ 4
r
+ 4 = (
r
+ 2)
2
= 0, which has the repeated root
r
1
=
r
2
=
r
=

2.
Then the general solution is
y
(
x
) =
c
1
e

2
x
+
c
2
xe

2
x
,
c
1
, c
2
= constant.
Since the I.C. are
given at
x
=

1, we can write the general solution as
y
(
x
) =
c
1
e

2(
x
+1)
+
c
2
xe

2(
x
+1)
.
Then
y
0
(
x
) =

2
c
1
e

2(
x
+1)
+ 2
c
2
e

2(
x
+1)

2
c
2
xe

2(
x
+1)
and hence
c
1

c
2
= 2 and

c
1
+ 3
c
2
= 1 which
yields
c
1
= 7 and
c
2
= 5.
Thus the general solution is
y
(
x
) = 7
e

2
x
+ 5
xe

2
x
, and decaying
exponentially as
x
→ ∞
.
b)
C.E. is 9
r
2

12
r
+ 4 = 0 and its roots are
r
1
=
r
2
= 2
/
3. Therefore, the general solution of the
D.E. is
y
=
c
1
e
2
x/
3
+
c
2
xe
2
x/
3
,
c
1
, c
2
= constant
The initial conditions yield
c
1
= 2 and
c
2
=

7
/
3. Since
r >
0,
y
→ ∞
as
x
→ ∞
.
c)
C.E. is 9
r
2
+ 6
r
+ 82 = 0 and its roots are
r
1
,
2
=

(1
/
3)
±
3
i
. Therefore, the general solution of
the D.E. is
y
=
c
1
e

x/
3
cos 3
x
+
c
2
e

x/
3
sin 3
x,
c
1
, c
2
= constant
The initial conditions yield
c
1
=

1 and
c
2
= 5
/
9.
Since the real part of the roots are negative,
y
→
0 as
x
→ ∞
.
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 '10
 YosumHoca
 SEPTA Regional Rail, #, Jaguar Racing, general solution, x+1, BILKENT UNIVERSITY Department of Mathematics MATH

Click to edit the document details