4. The problem is
y
′′

y
′
+
λy
= 0
,
0
< x <
2
,
y
′
(0) = 0
,
y
(2)

y
′
(2) = 0
.
This problem does nt look like an SL problem at first glance, but it might be rewritten so as to be an SL
problem. Is there a function
p
(
x
) such that
p
(
x
)
>
0 in [0
,
2] and multiplying the differential equation by
p
(
x
) it is of the form (
py
′
)
′
+
qy
+
λσy
= 0 for some
q, σ
?. We want these two equations to be the same, for
some function
p
:
0
=
p
(
y
′′

y
′
+
λy
) =
py
′′

py
′
+
λpy,
0
=
(
py
′
)
′
+
qy
+
λσy
=
py
′′
+
p
′
y
′
+
qy
+
λσy.
Clearly (I hope) we need (and it suﬃces)
p
such that
p
′
=

p
.
Solving this differential equation, we get
p
(
x
) =
Ce
−
x
, for constant
C
. We can take
C
= 1. We conclude: The differential equation is equivalent to
d
dx
(
e
−
x
y
)
+
λe
−
x
y
= 0
,
so that we have
a regular SL problem, with
p
(
x
) =
e
−
x
,
q
(
x
) = 0
,
σ
(
x
) =
e
−
x
,
κ
1
= 0
,
κ
2
= 1
,
κ
3
= 1
,
κ
4
=

1
.
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2
5. The problem is
2
y
′′
+
y
′
+ (
λ
+
x
)
y
= 0
,
0
< x <
∞
,
y
(0) = 0
,
y
(
x
)
, y
′
(
x
) bounded as
x
→ ∞
.
Solution.
The only question here is whether this is an SL problem; if it is, it is
singular
because the
interval is infinite among other reasons. To see if it is SL, we must see if there is a function
μ
(
x
) such that
after multiplying the differential equation by
μ
(
x
), the equation has the form (
py
′
)
′
+
qy
+
λσy
= 0. In ur
case, multiplication by a function
μ
(
x
) gives
0 = 2
μy
′′
+
μy
′
+
μxy
+
λμy.
We want this to be
0 =
py
′′
+
p
′
y
+
qy
+
λσy
= 0
.
Thus 2
μ
=
p
,
μ
=
p
′
,
q
=
μx
,
σ
=
μ
. From the first two equations,
p
′
=
1
2
p
; a non zero solution of this ODE
is
p
(
x
) =
e
x/
2
. Then
μ
=
1
2
e
x/
2
. The equation can be rewritten in the form
d
dx
(
e
x/
2
y
′
)
+
1
2
xe
x/
2
y
+
1
2
xe
x/
2
λy
= 0
,
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 Spring '13
 Schonbek
 Boundary value problem, Sturm–Liouville theory, SL problem, qy + λσy

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