MATH 215/255
Fall 2014
Assignment 5
§
2.5,
§
2.6
Solutions to selected exercises can be found in [Lebl], starting from page 303.
•
2.5.7:
a) Find a particular solution of
y
00

2
y
0
+
y
=
e
x
using the method of variation
of parameters.
b) Find a particular solution using the method of undetermined coefficients.
The second equation is satisfied for
u
2
=
x
, and the first then becomes
u
0
1
=

x
,
which is satisfied for
u
1
=

x
2
/
2. Our particular solution is now
y
p
=

1
2
x
2
e
x
+
x
2
e
x
=
1
2
x
2
e
x
.
Recalling the complementary solution (1), we have found the general form of the
solution:
y
=
C
1
e
x
+
C
2
xe

x
+
1
2
x
2
e
x
.
b) The righthand side is an exponential
e
x
, so we try first
y
p
=
Ae
x
. This fails quite
dramatically however, since we always obtain
y
00
p

2
y
0
p
+
y
p
= 0. A similar experiment
with
y
=
Axe
x
would yield the same result. We try finally
y
p
=
Ax
2
e
x
, for which
y
0
p
= (
Ax
2
+ 2
Ax
)
e
x
,
y
00
p
= (
Ax
2
+ 4
Ax
+ 2
A
)
e
x
.
Then
y
00
p

2
y
0
p
+
y
p
=
h
(
Ax
2
+ 4
Ax
+ 2
A
)

2(
Ax
2
+ 2
Ax
) +
Ax
2
i
e
x
= 2
Ae
x
.
We take therefore
A
= 1
/
2, so that after adding the complementary part (1), we find
the same general solution as before:
y
=
C
1
e
x
+
C
2
xe

x
+
1
2
x
2
e
x
.
•
2.5.9
For an arbitrary constant
c
find a particular solution to
y
00

y
=
e
cx
. Make
sure to handle every possible real
c
.