The equation below has a regular singular point at
x
0
= 0.
x
2
y
xy
(
x
2
1)
y
0
(a) Obtain the indicial equation for the ODE at x
0
= 0 and its two roots. (b) Then use all the information available and
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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 Fall '08
 STAFF
 Vector Space, Laplace, Regular singular point

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