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Mathematics 238 test two solutions
Tuesday, May 16, 2006
1
. State the
t
interval on which a solution to the following diﬀerential equation is guaranteed
to exist:
(4

t
2
)
y
00
+ 2
ty
0
+ 6
y
= ln
t
and
y
(1) = 0 and
y
0
(1) = 0
Rewrite the equation so that the coeﬃcient of the highest order derivative is equal to one:
y
00
+
±
2
t
4

t
2
²
y
0
+
±
6
4

t
2
²
y
=
ln
t
4

t
2
The domain on which all the coeﬃcients are continuous is 0
< t <
2 or
t >
2. The solution
to the initial condition involving
t
= 1 will exist inside this domain, so the solution will be
guaranteed to exist when 0
< t <
2.
2
. Circle each linearly independent set of functions:
(
a
)
{
f
(
t
) =
t
2

4
,g
(
t
) =
t
3

4
t
}
linearly independent set since one
function is not a
constant
multi
ple of the other
(
b
)
{
f
(
t
) = sin(2
t
)
,g
(
t
) = sin(

2
t
)
}
linearly
dependent
set
since
g
(
t
) = (

1)
f
(
t
)
(
c
)
{
F
(
t
) =
e
2
t
,G
(
t
) =
e

2
t
}
linearly independent set since
one function is not a constant
multiple of the other
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