Math 293 Solutions to Problem Set 10.
§
10.2 #4
.
y
00
+ 9
y
= 0 has general solution
y
=
c
1
sin 3
x
+
c
2
cos 3
x
.
Plugging in the
condition
y
(0) = 0 gives
c
2
= 0, so
y
=
c
1
sin 3
x
. To apply the condition
y
0
(
π
) =

6 we
compute
y
0
= 3
c
1
cos 3
x
, and then plugging in
y
0
(
π
) =

6 yields
c
1
= 2. The answer is
therefore
y
= 2 sin 3
x
.
#9
. To solve
y
00
+
λy
= 0 we examine the 3 possible cases:
Case (1):
λ >
0. Write
λ
=
α
2
with
α >
0. The quadratic equation is
r
2
+
α
2
= 0, with
roots
r
=
±
α i
, so the general solution is
y
=
c
1
sin
αx
+
c
2
cos
αx
. The boundary condition
y
(0) = 0 translates into
c
2
= 0, so we have
y
=
c
1
sin
αx
. The derivative is
y
0
=
αc
1
cos
αx
,
so the boundary condition
y
0
(
π
) = 0 implies that cos
απ
= 0.
(We are assuming that
α >
0, and if
c
1
were 0 then
y
would be the trivial solution 0.) The equation cos
απ
= 0
implies that
απ
=
2
n

1
2
π
for some positive integer
n
, hence
α
=
2
n

1
2
. Thus we have the
solutions
y
=
c
1
sin
2
n

1
2
x
corresponding to the eigenvalues
λ
=
α
2
=
(
2
n

1
2
)
2
.
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 Spring '07
 TERRELL,R
 Math, Differential Equations, Equations, Quadratic equation, Elementary algebra, Boundary value problem, R T

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