This preview shows page 1. Sign up to view the full content.
Math 415
Quiz 5A
Assuming the functions
y
1
and
y
2
are linearly independent solutions of
y
00
+
p
(
t
)
y
0
+
q
(
t
)
y
= 0 and that
y
3
=
y
1
+
y
2
and
y
4
=
y
1

y
2
, show that
y
3
and
y
4
also form a linearly independent set of solutions.
We know that the Wronskian of
y
1
and
y
2
,
W
(
y
1
,y
2
)(
t
) is nonzero on the interval on which the solution is
guaranteed to exist. So let’s consider the Wronskian of
y
3
and
y
4
.
W
(
y
3
,y
4
) =
±
±
±
±
±
y
1
+
y
2
y
1

y
2
y
0
1
+
y
0
2
y
0
1

y
0
2
±
±
±
±
±
= (
y
1
+
y
2
)(
y
0
1

y
0
2
)

(
y
1

y
2
)(
y
0
1
+
y
0
2
)
= (
y
1
y
0
1

y
1
y
0
2
+
y
2
y
0
1

y
2
y
0
2
)

(
y
1
y
0
1
+
y
1
y
0
2

y
2
y
0
1

y
2
y
0
2
) =

2
y
1
y
0
2
+ 2
y
2
y
0
1
=

2
W
(
y
1
,y
2
)
6
= 0
Hence,
y
3
and
y
4
are linearly independent.
Find the solution of the initial value problem:
y
00

4
y
0
+ 8
y
= 0
,
y
(
π/
2) = 0
,
y
0
(
π/
2) =

2
.
Describe its behavior as
t
increases.
The characteristic equation is
r
2

4
r
+ 8, which has roots
r
= 2
±
2
i
. So the general solution is
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: y ( t ) = C 1 e 2 t cos(2 t )+ C 2 e 2 t sin(2 t ). Plugging in the initial values tells us thatC 1 e π = 0 and2 C 1 e π2 C 2 e π =2. So C 1 = 0 and C 2 = eπ , and our solution is y ( t ) = e 2 tπ sin(2 t ). As for the behavior as t → ∞ , e 2 tπ grows without bound, and sin(2 t ) oscillates between positive and negative one, so the function oscillates without bound, getting closer and closer to both positive and negative inﬁnity by turns. 1...
View
Full
Document