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HW 8
1
1. Sec. 3.8:1. We have
y
00
+
&y
= 0
; y
0
(0) = 0
; y
(1) = 0
:
If
&
= 0
then we
have
y
00
= 0 =
)
y
0
=
a
=
)
y
(
x
) =
ax
+
b:
Now we use the endpoint conditions:
y
0
(0) =
a
= 0 =
)
a
= 0
;
y
(1) =
a
+
b
= 0 =
)
b
= 0
:
Thus, if
&
= 0 =
)
y
(
x
) = 0 =
)
&
= 0
is not eigenvalue. If
& >
0 =
)
&
=
±
2
>
0
;
so we have
y
00
+
±
2
y
= 0
:
The characteristic equation is
r
2
+
±
2
= 0 =
)
r
=
&
±i:
The solution has the following form
y
(
x
) =
A
cos
±x
+
B
sin
±x;
then
y
0
(
x
) =
±
A±
sin
±x
+
B±
cos
±x:
Now we use the endpoint condi
tions:
y
0
(0) =
= 0 =
)
B
= 0;
y
(1) =
A
cos
±
+
B
sin
±
= 0
=
)
A
cos
±
= 0 =
)
cos
±
= 0
:
Thus
±
=
&
2
+
²n
=
&
2
(1 + 2
n
)
; n
²
0;
then
&
n
=
±
2
=
²
2
4
(1 + 2
n
)
2
; n
²
0;
and the associated functions are
y
n
(
x
) = cos
²x
2
(1 + 2
n
)
; n
²
0
:
2. Sec. 3.8:3. We have
y
00
+
&y
= 0
; y
(
±
²
) = 0
; y
(
²
) = 0
:
If
&
= 0
then we
y
00
= 0 =
)
y
0
=
a
=
)
y
(
x
) =
ax
+
b
. Now we use the endpoint
conditions:
y
(
±
²
) =
±
a²
+
b
= 0
;
y
(
²
) =
a²
+
b
= 0
:
1
Note to the grader.
Please grade the following problems:
Sec. 3.8:1, Sec. 3.8:5, Sec. 3.8:8, Sec. 9.1: 3, Sec. 9.1:13
(20 pts for each problem)
1
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View Full DocumentSo,
2
b
= 0 =
)
b
= 0 =
)
a
= 0
:
Thus, if
&
= 0 =
)
y
(
x
) = 0 =
)
&
= 0
is not eigenvalue. If
& >
0 =
)
&
=
±
2
>
0
;
then we have
y
00
+
±
2
y
= 0
:
The characteristic equation is
r
2
+
±
2
= 0 =
)
r
=
&
±i:
The solution
has the following form
y
(
x
) =
A
cos
±x
+
B
sin
±x:
Now we use the endpoint conditions:
y
(
±
²
) =
A
cos
±²
±
B
sin
±²
= 0
;
y
(
²
) =
A
cos
±²
+
B
sin
±²
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 Fall '08
 Staff
 Differential Equations, Equations

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