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
Thus, the indicial equation is
r
2
 2
r
 8 = 0, with two roots
r
1
= 4 and
r
2
= 2.
Consequently the two linearly independent solutions provided by
Theorem 6.3 look like the following:
and
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentTEST3/MAP2302
Page 2 of 4
______________________________________________________________________
3. (10 pts.)
Obtain the general solution to the following linear ODE:
By letting
x
= e
t
, and
w
(
t
) =
y
(e
t
), so that
y
(
x
) =
w
(ln(
x
)) for
x
> 0, the ODE above transforms into the following ODE in
w
(
t
):
w
(
t
)
4
w
(
t
)
2
t
.
The corresponding homogeneous equ.:
w
(
t
)
4
w
(
t
)
0.
The auxiliary equation:
m
2
+ 4 = (
m
+ 2
i
)(
m
 2
i
) = 0
Here’s a fundamental set of solutions for the corresponding homogeneous
equation:
{ cos(2
t
),sin(2
t
) }
The driving function of the transformed equation is a U.C. function.
By muttering the appropriate incantation and waving your magic writing
utensil over the exam, you find that
w
p
(
t
)
1
2
t
is a particular integral. Consequently, the general solution to the
original ODE, the one involving
y
, is
y
(
x
)
c
1
cos(2 ln(
x
))
c
2
sin(2 ln(
x
))
1
2
ln(
x
)
for x
> 0.
______________________________________________________________________
4. (15 pts.)
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 STAFF
 Derivative, #, 5 pts, 10 pts, 4 W

Click to edit the document details