AM 351: Assignment 4: Solutions
1.
a) We look for a solution of the form
x
r
and obtain the characteristic equation,
r
2
+ (
α

1)
r
+
β
= 0
.
The roots of which are,
r
=
1

α
±
p
(
α

1)
2

4
β
2
.
If we only have one root then that requires (
α

1)
2
= 4
β
. But from this substitute we
ﬁnd one solution,
y
1
(
x
) =
x
r
1
,
where
r
1
= (1

α
)
/
2 or
α
= 1

2
r
1
.
We look for a second solution of the form
y
2
=
v
(
x
)
y
1
and know, from our previous work
that
v
is determined by the following equation,
v
0
=
1
x
2
r
1
exp
±

Z
α
x
dx
²
=
1
x
2
r
1
exp (

α
ln
x
) =
x
2
r
1

1
x
2
r
1
=
1
x
.
Therefore,
v
= ln
x
and
y
2
=
x
r
1
ln
x
b) From assignment 1 recall that if we substitute
x
=
e
t
then we get
dy
dx
=
e

t
dy
dt
,
d
2
y
dx
2
=
e

2
t
³
d
2
y
dt
2

dy
dt
´
.
Similarly, we obtain for the third order derivative,
d
3
y
dx
3
=
e

3
t
³
d
3
y
dt
3

3
d
2
y
dt
2
+ 2
dy
dt
´
When we substitute this into the equation we get,
d
3
y
dt
3

3
d
2
y
dt
2
+ 3
dy
dt

y
= 0
.
If we try
y
=
e
λt
then the characteristic equation is,
λ
3

3
λ
2
+ 3
λ

1 = 0
This is a cubic equation that can be very diﬃcult to solve but observe that
λ
= 1 is a
valid root. Then by long division we get the other two and
(
λ

1)
3
= 0
Therefore the roots are
λ
= 1 with multiplicity of 3. We know that the general solution
to this equation should be,
y
(
t
) =
c
1
e
t
+
c
2
te
t
+
c
3
t
2
e
t
,
and when we substitute back in terms of
x
we get,
y
(
x
) =
c
1
x
+
c
2
(ln
x
)
x
+
c
3
(ln
x
)
2
x.
1
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a) We compute
p
0
and
q
0
, as deﬁned in lecture,
p
0
= lim
x
→
1
(
x

1)
±

2
x
1

x
2
²
=
2
2
= 1
,
q
0
= lim
x
→
1
(
x

1)
2
±
p
(
p
+ 1)
1

x
2
²
= 0
.
So the associated Euler equation is,
x
2
y
00
+
xy
0
= 0
.
The characteristic equation is
r
(
r

1) +
r
= 0 that implies
r
= 0 is a double root.
b) Before we look for a power series solution I choose to substitute
t
=
x

1. If we rewrite
Y
(
t
) =
y
(
x
) then the equation is,
(

t
2

2
t
)
Y
00
+ (

2
t

2)
Y
0
+
p
(
p
+ 1)
Y
= 0
.
Therefore, we look for a regular power series solution of the form,
Y
=
∞
X
n
=0
a
n
t
n
We substitute this into the equation and rewrite each term separately.
p
(
p
+ 1)
Y
=
∞
X
n
=0
p
(
p
+ 1)
a
n
t
n
,

2
tY
0
=
∞
X
n
=0
(

2
n
)
a
n
t
n
,

2
Y
0
=
∞
X
n
=1
(

2
n
)
a
n
t
n

1
=
∞
X
n
=0
(

2(
n
+ 1))
a
n
+1
t
n
,

t
2
Y
00
=
∞
X
n
=0

n
(
n

1)
a
n
t
n
,

2
tY
00
=
∞
X
n
=0

2
n
(
n

1)
a
n
t
n

1
=
∞
X
n
=0

2(
n
+ 1)(
n
)
a
n
+1
t
n
.
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 Spring '14
 Trigraph, Characteristic polynomial, dx, Frobenius method

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