MA 2930, March 30, 2011
Worksheet 9
1.
Find the solution (if possible) of
y
00
+
y
= 0
for the following sets of boundary
values:
(a)
y
(0) = 0
,
y
(
π
) = 0
(b)
y
(0) = 0
,
y
0
(
π
) = 0
(c)
y
0
(0) = 0
,
y
(
π
) = 0
(d)
y
0
(0) = 0
,
y
0
(
π
) = 0
How would you
make sense
of these results? (Hint: think graphically!)
For which of these boundary values would the equation
y
00
+
y
= cos 2
x
be
solvable?
The general solution of the differential equation is
y
(
x
) =
c
1
cos
x
+
c
2
sin
x
Now let’s apply the boundary conditions:
(a)
y
(0) =
c
1
= 0 and
y
(
π
) =

c
1
= 0.
So the solutions are
y
(
x
) =
c
2
sin
x
. Graphically this makes sense because any sine with period 2
π
can
satisfy the boundary conditions, but no cosine. Note that there are infinitely
many solutions, illustrating that there is no uniqueness result for BVPs.
(b)
y
(0) =
c
1
= 0 and
y
0
(
π
) =

c
2
= 0. So the only solution is
y
(
x
) = 0.
(c)
y
0
(0) =
c
2
= 0 and
y
(
π
) =

c
1
= 0. So the only solution is again
y
(
x
) = 0.
(d)
y
0
(0) =
c
2
= 0 and
y
0
(
π
) =

c
2
= 0. So the solutions are
y
(
x
) =
c
1
cos
x
.
The general solution of the differential equation of
y
00
+
y
= cos 2
x
is
y
(
x
) =
c
1
cos
x
+
c
2
sin
x

1
/
3 cos 2
x
Now let’s apply the boundary conditions:
(a)
y
(0) =
c
1

1
/
3 = 0 and
y
(
π
) =

c
1

1
/
3 = 0.
So there is no
solution. Graphically this makes sense because no cosine can satisfy these
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 Spring '07
 TERRELL,R
 Differential Equations, Equations, Fourier Series, Sets, Sin, Cos, Periodic function, α

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