Theorem 6.3 to say what the two nontrivial linearly independent solutions given by theorem look like without attempting to
obtain the coefficients of the power series involved.
To determine
r
, you need the indicial equation at x
0
= 0
and its
roots.
Now the indicial equation is
r
(
r
1) +
p
0
r
+
q
0
= 0 where
p
0
lim
x
→
0
x
x
x
2
1 and
q
0
lim
x
→
0
x
2
x
2
1
x
2
1.
Thus, the indicial equation is
r
2
 2
r
+ 1 = 0, with a single root
r
1
= 1
with multiplicity two.
Consequently the two linearly independent solutions
provided by Theorem 6.3 look like the following:
and
y
1
(
x
)
x
1
∞
n
0
c
n
x
n
y
2
(
x
)
x
2
∞
n
0
d
n
x
n
y
1
(
x
)ln
x
______________________________________________________________________
6.
(10 pts.)
Compute
when
f
(
t
)
1
{
F
(
s
)}(
t
)
(a)
F
(
s
)
7
(2
s
1)
3
(
Work
)
7
8(
s
(
1/2))
3
1
{
F
(
s
)}(
t
)
7
8
1
2
t
2
e
(1/2)
t
7
16
t
2
e
(1/2)
t
(b)
F
(
s
)
2
s
12
s
2
6
s
13
(
Work
)
2
s
12
(
s
3)
2
4
2(
s
3)
6
(
s
3)
2
2
2
1
{
F
(
s
)}(
t
)
2
e
3
t
cos(2
t
)
3
e
3
t
sin(2
t
)
’Tis the usual prestidigitation of multiplication by ’1’ in the correct
form or the addition of ’0’ suitably transmogrified.
I move that we table
the motion.
Do I hear a second?
______________________________________________________________________
7.
(5 pts.)
Circle the letter corresponding to the correct
response: If
,then
F
(
s
) =
F
(
s
)
{
t
2
sin(
bt
)}(
s
)
(a)
(b)
2
s
3
b
(
s
2
b
2
)
d
ds
2
bs
(
s
2
b
2
)
2
(c)
(d)
d
2
ds
2
b
s
2
b
2
2
bs
s
2
(
s
2
b
2
)
2
(e)
None of the above.
Obviously (a) and (d) are utter nonsense since the transform of a product
is NOT the product of the transforms.
(c) is a near miss since the sign is
wrong.
It turns out that (b) provides the correct answer.
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______________________________________________________________________
8.
(15 pts.)
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 Fall '08
 STAFF
 Vector Space, Laplace, Regular singular point

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