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
→ ∞
.
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