Series sol. near ordinary points w/ given initial conds.
Let’s revisit the 2 problems that we have solved in class – but let’s solve them with some
initial conditions. We will solve them two using two different methods.
1. Using the results we obtained in
ordinary_ps.pdf
.
2. Using a Taylor series expansion of the solution about
x
0
.
Before doing so, recall that the Talor series expansion of a function about point
x
0
is given
by
y
(
x
) =
y
(
x
0
) +
y
0
(
x
0
)(
x

x
0
) +
y
00
(
x
0
)
2!
(
x

x
0
)
2
+
y
000
(
x
0
)
3!
(
x

x
0
)
3
+
y
(4)
(
x
0
)
4!
(
x

x
0
)
4
+
· · ·
.
(1)
For a given second order initial value problem (IVP), we already know
y
(
x
0
)
and
y
0
(
x
0
)
.
What we need to find are
y
00
(
x
0
)
, y
000
(
x
0
)
,
· · ·
. We will find these using the given differential
equation, as illustrated in the problems below.
1
c HF, 2011
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Series sol. near ordinary points w/ given initial conds.
Problem 1
(1 +
x
)
y
00
+
y
= 0
,
x
0
= 0
.
y
(0) = 1
, y
0
(0) = 1
.
(2)
Solution – Method A
:
Recall that we previous found the general solution to the problem to be (see
ordinary_ps.pdf
)
y
=
a
0
1

1
2
x
2
+
1
6
x
3

1
24
x
4
+
· · ·
+
a
1
x

1
6
x
3
+
1
12
x
4
+
· · ·
,
(3)
where
a
0
and
a
1
are two arbitrary constants. Using the first initial conditions, we find
y
(0) = 1 =
a
0
1

1
2
0
2
+
1
6
0
3

1
24
0
4
+
· · ·
+
a
1
0

1
6
0
3
+
1
12
0
4
+
· · ·
⇒
a
0
= 1
.
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 Spring '10
 None
 Derivative, Taylor Series, Constant of integration, Boundary value problem

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