Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

22.5 Homogeneous Constant Coefficient Equations
Homogeneous constant coefficient equations have the form
a
n+N
+ p
N−1
a
n+N−1
+ ··· + p
1
a
n+1
+ p
0
a
n
= 0.
The substitution a
n
= r
n
yields
r
N
+ p
N−1
r
N−1
+ ··· + p
1

r + p
0
= 0
(r −r
1
)
m
1
···(r −r
k
)
m
k
= 0.
If r
1
is a distinct root then the associated linearly independent solution is r
n
1
. If r
1
is a root of multiplicity m > 1
then the associated solutions are r
n
1
, nr
n
1
, n
2

r
n
1
, . . . , n
m−1
r
n
1
.
Result 22.5.1 Consider the homogeneous constant coefficient difference equation
a
n+N
+ p
N−1
a
n+N−1
+ ··· + p
1
a
n+1
+ p
0
a
n
= 0.
The substitution a
n
= r
n
yields the equation

(r −r
1
)
m
1
···(r − r
k
)
m
k
= 0.
A set of linearly independent solutions is
{r
n
1
, nr
n
1
, . . . , n
m
1
−1
r
n
1
, . . . , r
n
k
, nr
n

k
, . . . , n
m
k
−1
r
n
k
}.
Example 22.5.1 Consider the equation a
n+2
− 3a
n+1
+ 2a
n
= 0 with the initial conditions a
1
= 1 and a
2
= 3. The
substitution a
n
= r
n
yields
r
2
− 3r + 2 = (r −1)(r − 2) = 0.
Thus the general solution is
a

n
= c
1
1
n
+ c
2
2
n
.
1174
The initial conditions give the two equations,
a
1
= 1 = c
1
+ 2c
2
a
2
= 3 = c
1
+ 4c
2
Since c
1
= −1 and c
2
= 1, the solution to the difference equation subject to the initial conditions is
a

n
= 2
n
− 1.
Example 22.5.2 Consider the gambler’s ruin problem that was introduced in Example 22.1.1. The equation for the
probabili ty of the gambler’s ruin at n dollars is
a
n
= pa
n+1
+ qa
n−1
subject to a
0
= 1, a
N
= 0.
We assume that 0 < p < 1. With the substitution a
n
= r
n
we obtain
r = pr
2
+ q.
The roots of this equation are
r =
1 ±
√
1 − 4pq

2p
=
1 ±

1 − 4p(1 − p)
2p
=
1 ±

(1 − 2p)
2
2p
=
1 ± |1 − 2p|
2p
.
We will consider the two cases p = 1/2 and p = 1/2.
1175
p = 1/2. If p < 1/2, the ro ots are
r =
1 ± (1 − 2p)
2p
r
1
=
1 − p
p
=
q
p

, r
2
= 1.
If p > 1/2 the roots are
r =
1 ± (2p − 1)
2p
r
1
= 1, r
2
=
−p + 1
p
=
q
p
.
Thus the general solution for p = 1/2 is
a
n
= c
1
+ c
2

q
p

n

.
The boundary condition a
0
= 1 requires that c
1
+ c
2
= 1. From the boundary condition a
N
= 0 we have
(1 − c
2
) + c
2

q
p

N
= 0
c
2
=
−1
−1 + (q/p)
N
c
2
=
p

N
p
N
− q
N
.
Solving for c
1
,
c
1
= 1 −
p
N
p
N
− q
N
c
1
=
−q
N
p
N
− q
N
.
1176
Thus we have

a
n
=
−q
N
p
N
− q
N
+
p
N
p
N
− q
N

q
p

n
.
p = 1/2. In this case, the two roots of the polynomial are both 1. The general solution is
a
n
= c
1
+ c
2
n.

The left boundary condition demands that c
1
= 1. From the right boundary condition we obtain
1 + c
2
N = 0
c
2
= −
1
N
.
Thus the solution for this case is
a
n
= 1 −
n
N
.
As a check that this formula makes sense, we see that for n = N/2 the probability of ruin is 1 −
N/2
N
=
1
2
.
22.6 Reduction of Order
Consider the difference equation
(n + 1)(n + 2)a
n+2

− 3(n + 1)a
n+1
+ 2a
n
= 0 for n ≥ 0 (22.1)
We see that one solution to this equation is a
n
= 1/n!. Analogous to the reduction of order for differential equations,
the substitution a
n
= b
n
/n! will reduce the order of the difference equation.
(n + 1)(n + 2)b
n+2
(n + 2)!
−
3(n + 1)b
n+1
(n + 1)!
+
2b
n
n!
= 0
b
n+2
− 3b
n+1
+ 2b

n
= 0 (22.2)
1177
At first glance it appears th at we have not reduced the order of the equation, but writing it in terms of discrete
derivatives
D
2
b
n
− Db
n
= 0
shows that we now have a first order difference e quation for Db
n
. The substitution b
n
= r
n
in equation 22.2 yields the
algebraic equation
r
2
− 3r + 2 = (r −1)(r − 2) = 0.
Thus the solutions are b
n
= 1 and b
n
= 2
n
. Only the b

n
= 2
n
solution will give us another linearly independent solution
for a
n
. Thus the second solution for a
n
is a
n
= b
n
/n! = 2
n
/n!. The general solution to equation 22.1 is then
a
n
= c
1
1
n!
+ c
2
2
n
n!
.
Result 22.6.1 Let a
n
= s

n
be a homogeneous solution of a linear difference equation. The
substitution a
n
= s
n
b
n
will yield a difference equation for b
n
that is of order one less than the
equation for a
n
.
1178
22.7 Exercises
Exercise 22.1
Find a formula for the n
th
term in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, . .
Hint, Solution
Exercise 22.2
Solve the difference equation
a
n+2
=
2
n
a
n

, a
1
= a
2
= 1.
Hint, Solution
1179
22.8 Hints
Hint 22.1
The difference equation corresponding to the Fibonacci sequence is
a
n+2
− a
n+1
− a
n
= 0, a
1
= a
2
= 1.
Hint 22.2
Consider this exercise as two first order difference equations; one for the even terms, one for the odd terms.
1180
22.9 Solutions
Solution 22.1
We can describe the Fibonacci sequence with the difference equation
a
n+2
− a

n+1
− a
n
= 0, a
1
= a
2
= 1.
With the substitution a
n
= r
n
we obtain the equation
r
2
− r −1 = 0.
This equation has the two distinct roots
r
1
=
1 +
√
5
2
, r
2
=
1 −
√
5

2
.
Thus the general solution is
a
n
= c
1

1 +
√
5
2

n
+ c
2

1 −
√
5
2

n
.
From the initial conditions we have
c
1
r
1
+c

2
r
2
= 1
c
1
r
2
1
+c
2
r
2
2
= 1.
Solving for c
2
in the first equation,
c
2
=
1
r
2
(1 − c
1
r
1
).
We subs titute this into the second equation.

c
1
r
2
1
+
1
r
2
(1 − c
1
r
1
)r
2
2
= 1
c
1
(r
2
1
− r
1
r
2
) = 1 − r
2
1181
c

1
=
1 − r
2
r
2
1
− r
1
r
2
=
1 −
1−
√
5
2
1+
√
5
2
√
5
=
1+
√
5
2
1+
√

5
2
√
5
=
1
√
5
Substitute this result into the equation for c
2
.
c
2
=
1
r
2

1 −
1
√
5
r
1

=
2
1 −
√
5


1 −
1
√
5
1 +
√
5
2

= −
2
1 −
√
5

1 −
√
5
2
√
5

= −
1
√
5
Thus the n
th
term in the Fibonacci sequence has the formula

a
n
=
1
√
5

1 +
√
5
2

n
−
1
√
5

1 −
√
5
2

n
.
It is interesting to note that although the Fibonacci sequence is defined in terms of integers, one cannot express the
formula form the n
th
element in terms of rational numbers.
1182

Solution 22.2
We can consider
a
n+2
=
2
n
a
n
, a
1
= a
2
= 1
to be a first order difference equation. First consider the o dd terms.
a
1
= 1
a
3
=
2
1
a
5
=
2
3
2
1

a
n
=
2
(n−1)/2
(n − 2)(n − 4) ···(1)
For the even terms,
a
2
= 1
a
4
=
2
2
a
6
=
2
4
2
2
a
n
=
2
(n−2)/2
(n − 2)(n − 4) ···(2)
.
Thus

a
n
=

2
(n−1)/2
(n−2)(n−4)···(1)
for odd n
2
(n−2)/2
(n−2)(n−4)···(2)
for even n.
1183
Chapter 23
Series Solutions of Differential Equations
Skill beats honesty any day.
-?
23.1 Ordinary Points
Big O and Little o Notation. The notation O(z
n
) means “terms no bigger than z
n
.” This gives us a convenient
shorthand for manipulating series. For example,
sin z = z −
z
3
6
+ O(z
5

)
1
1 − z
= 1 + O(z)
The notation o(z
n
) means “terms smaller that z
n
.” For example,
cos z = 1 + o(1)
e
z
= 1 + z + o(z)
1184
Example 23.1.1 Consider the equation
w

(z) − 3w

(z) + 2w(z) = 0.
The general solution to this constant coefficient equation is
w = c
1
e
z
+c
2
e
2z
.

The functions
e
z
and
e
2z
are analytic in the finite complex plane. Recall that a function is analytic at a point z
0
if and
only if the function has a Taylor series about z
0
with a nonzero radius of convergence. If we substitute the Taylor series
expansions about z = 0 of
e
z
and
e
2z
into the general solution, we obtain
w = c
1
∞

n=0
z
n
n!
+ c
2
∞


n=0
2
n
z
n
n!
.
Thus we have a series solution of the differential equation.
Alternatively, we could try substituting a Taylor series into the differential equation and solving for the coefficients.
Substituting w =

∞
n=0
a
n
z
n
into the differential equation yields
d
2
dz
2
∞

n=0
a
n
z
n

− 3
d
dz
∞

n=0
a
n
z
n
+ 2
∞

n=0
a
n
z
n
= 0
∞

n=2
n(n − 1)a
n
z
n−2
− 3
∞

n=1

na
n
z
n−1
+ 2
∞

n=0
a
n
z
n
= 0
∞

n=0
(n + 2)(n + 1)a
n+2
z
n
− 3
∞

n=0
(n + 1)a
n+1
z
n
+ 2
∞


n=0
a
n
z
n
= 0
∞

n=0

(n + 2)(n + 1)a
n+2
− 3(n + 1)a
n+1
+ 2a
n

z
n
= 0.
Equating powers of z, we obtain the difference equation
(n + 2)(n + 1)a
n+2
− 3(n + 1)a
n+1
+ 2a
n
= 0, n ≥ 0.
1185

We see that a
n
= 1/n! is one solution since
(n + 2)(n + 1)
(n + 2)!
− 3
n + 1
(n + 1)!
+ 2
1
n!
=
1 − 3 + 2
n!
= 0.
We use reduction of order for difference equations to find th e other solution. Substituting a
n
= b
n
/n! into the difference
equation yields
(n + 2)(n + 1)
b
n+2
(n + 2)!
− 3(n + 1)
b
n+1
(n + 1)!
+ 2

b
n
n!
= 0
b
n+2
− 3b
n+1
+ 2b
n
= 0.
At first glance it appears that we have not reduced the order of the difference equation. However writing this equation
in terms of discrete derivatives,
D
2
b
n
− Db
n
= 0
we see that this is a first order difference equation for Db
n
. Since this is a constant coefficient difference equation we
substitute b
n
= r
n
into the equation to obtain an algebraic equation for r.
r
2

− 3r + 2 = (r −1)(r − 2) = 0
Thus the two solutions are b
n
= 1
n
b
0
and b
n
= 2
n
b
0
. Only b
n
= 2
n
b
0
will give us a second independent solution for a
n
.
Thus the two solutions for a
n
are
a
n
=
a
0

n!
and a
n
=
2
n
a
0
n!
.
Thus we can write the general solution to the differential equation as
w = c
1
∞

n=0
z
n
n!
+ c
2
∞

n=0
2
n
z
n
n!
.

We recognize these two sums as the Taylor expansions of
e
z
and
e
2z
. Thus we obtain the same result as we did solving
the differential equation directly.
1186
Of course it would be pretty silly to go through all the grunge i nvolved in developing a series expansion of the solution
in a problem like Example 23.1.1 since we can solve the problem exactly. However if we could not solve a differential
equation, then having a Taylor series expansion of the solution about a point z
0
would be useful in determining the
behavior of the solutions near that point.
For this method of substituting a Taylor series into the differential equation to be useful we have to know at what
points the solutions are analytic. Let’s say we were considering a second order differential equation whose solutions
were
w
1
=
1
z
, and w
2
= log z.
Trying to find a Taylor series expansion of the solutions about the point z = 0 would fail because the solutions are not
analytic at z = 0. This brings us to two important questions.
1. Can we tell if the solutions to a linear differential equation are analytic at a point without knowing the solutions?
2. If there are Taylor series expansions of the solutions to a differential equation, what are the radii of convergence

of the series?
In order to answer these questions, we will introduce the concept of an ordinary point. Consider the n
th
order linear
homogeneous equation
d
n
w
dz
n
+ p
n−1
(z)
d
n−1
w
dz
n−1
+ ··· + p
1
(z)
dw
dz
+ p
0
(z)w = 0.
If each of the coefficient functions p
i
(z) are analytic at z = z
0

then z
0
is an ordinary point of the differential equation.
For reasons of typography we will restrict our attention to second order equations and the p oint z
0
= 0 for a while.
The generalization to an n
th
order equation will be apparent. Considering the point z
0
= 0 is only trivially more general
as we could introduce the transformation z − z
0
→ z to move the point to the origin.
In the chapter on first order differential equations we showed that the solution is analytic at ordinary poin ts. One
would guess that this remains true for higher order equations. Consider the second order equation
y

+ p(z)y

+ q(z)y = 0,
where p and q are analytic at the origin.
p(z) =
∞

n=0
p
n
z
n

, and q(z) =
∞

n=0
q
n
z
n
1187
Assume that one of the solutions is not analytic at the origin and behaves like z
α
at z = 0 where α = 0, 1, 2, . . That
is, we can approximate the solution with w(z) = z
α
+ o(z
α
). Let’s substitute w = z
α
+ o(z
α
) into the differential
equation and look at the lowest power of z in each of the terms.

α(α − 1)z
α−2
+ o(z
α−2
)

+


αz
α−1
+ o(z
α−1
)

∞

n=0
p
n
z
n
+

z
α
+ o(z
α
)

∞

n=0
q
n
z
n
= 0.

We se e that the solution could not possibly behave like z
α
, α = 0, 1, 2, ··· because there is no term on the left to
cancel out the z
α−2
term. The terms on the left side could not add to zero.
You could also check that a solution could not possibly behave like log z at the origin. Though we will not prove
it, if z
0
is an ordinary point of a homogeneous differential equation, then all the solutions are analytic at the point z
0
.
Since the solution is analytic at z
0
we can expand it in a Taylor series.
Now we are prepared to answer our second question. From complex variables, we know that the radius of convergence
of the Taylor series expansion of a function is the distance to the nearest singularity of that function. Since the solutions
to a differential equation are analytic at ordinary points of the equation, the series expansion about an ordinary point
will have a radius of convergence at least as large as the distance to the nearest singularity of the coefficient functions.
Example 23.1.2 Consider the equation
w

+
1
cos z
w

+ z
2
w = 0.

If we expand the solution to the differential equation in Taylor series about z = 0, the radius of convergence will be at
least π/2. This is because the coefficient functions are analytic at the origin, and the nearest singularities of 1/ cos z
are at z = ±π/2.
23.1.1 Taylor Series Expansion for a Second Order Diff ere ntial Equation
Consider the differential equation
w

+ p(z)w

+ q(z)w = 0
1188
where p(z) and q(z) are analytic in some neighborhood of the origin.
p(z) =
∞

n=0
p
n
z
n
and q(z) =
∞

n=0
q
n
z
n
We subs titute a Taylor series and it’s derivatives
w =

∞

n=0
a
n
z
n
w

=
∞

n=1
nz
n
z
n−1
=
∞

n=0
(n + 1)a
n+1
z
n
w

=
∞


n=2
n(n − 1)a
n
z
n−2
=
∞

n=0
(n + 2)(n + 1)a
n+2
z
n
into the differential equation to obtain
∞

n=0
(n + 2)(n + 1)a
n+2
z
n
+

∞

n=0
p
n
z
n


∞

n=0
(n + 1)a
n+1
z
n

+

∞

n=0
q
n
z
n

∞

n=0
a
n
z
n

= 0
∞


n=0
(n + 2)(n + 1)a
n+2
z
n
+
∞

n=0

n

m=0
(m + 1)a
m+1
p
n−m

z
n
+
∞

n=0

n

m=0
a
m

q
n−m

z
n
= 0
∞

n=0

(n + 2)(n + 1)a
n+2
+
n

m=0

(m + 1)a
m+1
p
n−m
+ a
m
q
n−m


z
n
= 0.

1189
Equating coefficients of powers of z,
(n + 2)(n + 1)a
n+2
+
n

m=0

(m + 1)a
m+1
p
n−m
+ a
m
q
n−m

= 0 for n ≥ 0.
We see that a
0
and a
1
are arbitrary and the rest of the coefficients are determined by the recurrence relation
a
n+2
= −
1
(n + 1)(n + 2)
n


m=0
((m + 1)a
m+1
p
n−m
+ a
m
q
n−m
) for n ≥ 0.
Example 23.1.3 Consider the problem
y

+
1
cos x
y

+
e
x
y = 0, y(0) = y

(0) = 1.
Let’s expand the solution in a Taylor series about the origin.
y(x) =
∞

n=0

a
n
x
n
Since y(0) = a
0
and y

(0) = a
1
, we see that a
0
= a
1
= 1. The Taylor expansions of the coefficient functions are
1
cos x
= 1 + O(x), and
e
x
= 1 + O(x).
Now we can calculate a
2
from the recurrence relation.
a
2
= −
1
1 · 2
0


m=0
((m + 1)a
m+1
p
0−m
+ a
m
q
0−m
)
= −
1
2
(1 · 1 · 1 + 1 · 1)
= −1
1190
0.2 0.4 0.6 0.8
1
1.2
1.4
0.7
0.8
0.9
1.1
1.2
Figure 23.1: Plot of the Numerical Solution and the First Three Terms in the Taylor Series.
Thus the solution to the problem is
y(x) = 1 + x −x
2

+ O(x
3
).
In Figure 23.1 the numerical solution is plotted i n a solid line and the sum of the first three terms of the Taylor series
is plotted in a dashed line.
The general recurrence relation for the a
n
’s is useful if you only want to calculate the first few terms in the Taylor
expansion. However, for many problems substituting the Taylor series for the coefficient functions into the differential
equation will enable you to find a simpler form of the solution. We consider the foll owing example to illustrate this
point.
1191
Example 23.1.4 Develop a series expansion of the solution to the initial value problem
w

+
1
(z
2
+ 1)
w = 0, w(0) = 1, w

(0) = 0.
Solution using the General Recur re nce Relation. The coefficient function has the Taylor expansion
1
1 + z
2
=
∞


n=0
(−1)
n
z
2n
.
From the initial condition we obtain a
0
= 1 and a
1
= 0. Thus we see that the solution is
w =
∞

n=0
a
n
z
n
,
where
a
n+2
= −
1
(n + 1)(n + 2)
n

m=0
a

m
q
n−m
and
q
n
=

0 for odd n
(−1)
(n/2)
for even n.
Although this formula is fine if you only want to calculate the first few a
n
’s, it is just a tad unwieldy to work with.
Let’s see if we can get a better expression for the solution.
Substitute the Taylor Series into the Differential Equation. Substituting a Taylor series for w yields
d
2
dz
2
∞

n=0
a
n
z
n
+
1

(z
2
+ 1)
∞

n=0
a
n
z
n
= 0.
1192
Note that the algebra will be easier if we multiply by z
2
+ 1. The polynomial z
2
+ 1 has only two terms, but the Taylor
series for 1/(z
2
+ 1) has an infinite number of terms.
(z
2
+ 1)
d
2
dz
2
∞

n=0

a
n
z
n
+
∞

n=0
a
n
z
n
= 0
∞

n=2
n(n − 1)a
n
z
n
+
∞

n=2
n(n − 1)a
n
z
n−2
+
∞


n=0
a
n
z
n
= 0
∞

n=0
n(n − 1)a
n
z
n
+
∞

n=0
(n + 2)(n + 1)a
n+2
z
n
+
∞

n=0
a
n
z
n

= 0
∞

n=0

(n + 2)(n + 1)a
n+2
+ n(n − 1)a
n
+ a
n

z
n
= 0
Equating powers of z gives us the difference equation
a
n+2
= −
n
2
− n + 1
(n + 2)(n + 1)
a
n
, for n ≥ 0.
From the initial conditions we see that a
0
= 1 and a
1

= 0. All of the odd terms in the series will be zero. For the
even terms, it is easier to reformulate the problem with the change of variables b
n
= a
2n
. In terms of b
n
the difference
equation is
b
n+1
= −
(2n)
2
− 2n + 1
(2n + 2)(2n + 1)
b
n
, b
0
= 1.
This is a first order difference equation with the solution
b
n
=
n

j=0

−

4j
2
− 2j + 1
(2j + 2)(2j + 1)

.
Thus we have that
a
n
=


n/2
j=0

−
4j
2
−2j+1
(2j+2)(2j+1)

for even n,
0 for odd n.
1193
Note that the nearest singularities of 1/(z
2
+1) in the complex plane are at z = ±i. Thus the radius of convergence
must be at least 1. Applying the ratio test, the series converges for values of |z| such that
lim
n→∞





a
n+2
z
n+2
a
n
z
n




< 1
lim
n→∞




−
n
2
− n + 1
(n + 2)(n + 1)





|z|
2
< 1
|z|
2
< 1.
The radius of convergence is 1.
The first few terms in the Taylor expansion are
w = 1 −
1
2
z
2
+
1
8
z
4
−
13
240
z
6
+ ··· .
In Figure 23.2 the plot of the first two nonzero terms is shown in a short dashed line, the plot of the first four
nonzero terms is shown in a long dashed line, and the numerical solution is shown in a solid line.
In general, if the coefficient functions are rational functions, that is they are fractions of polynomials, multiplying
the equations by the quotient will reduce the algebra involved in finding the series solution.

Example 23.1.5 If we were going to find the Taylor series expansi on about z = 0 of the solution to
w

+
z
1 + z
w

+
1
1 − z
2
w = 0,
we would first want to multiply the equation by 1 −z
2
to obtain
(1 − z
2
)w

+ z(1 − z)w

+ w = 0.
Example 23.1.6 Find the series expansions about z = 0 of the fundamental set of solutions for
w

+ z
2
w = 0.
1194

0.2 0.4 0.6 0.8
1
1.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Figure 23.2: Plot of the solution and approximations.
Recall that the fundamental set of solutions {w
1
, w
2
} satisfy
w
1
(0) = 1 w
2
(0) = 0
w

1
(0) = 0 w

2
(0) = 1.
Thus if
w

1
=
∞

n=0
a
n
z
n
and w
2
=
∞

n=0
b
n
z
n
,
then the coefficients must satisfy
a
0
= 1, a
1
= 0, and b
0
= 0, b
1
= 1.

1195
Substituting the Taylor expansion w =

∞
n=0
c
n
z
n
into the differential equation,
∞

n=2
n(n − 1)c
n
z
n−2
+
∞

n=0
c
n
z
n+2
= 0
∞

n=0
(n + 2)(n + 1)c

n+2
z
n
+
∞

n=2
c
n−2
z
n
= 0
2c
2
+ 6c
3
z +
∞

n=2

(n + 2)(n + 1)c
n+2
+ c
n−2

z
n
= 0
Equating coefficients of powers of z,

z
0
: c
2
= 0
z
1
: c
3
= 0
z
n
: (n + 2)(n + 1)c
n+2
+ c
n−2
= 0, for n ≥ 2
c
n+4
= −
c
n
(n + 4)(n + 3)
For our first solution we have the difference equation
a
0
= 1, a
1
= 0, a
2

= 0, a
3
= 0, a
n+4
= −
a
n
(n + 4)(n + 3)
.
For our second solution,
b
0
= 0, b
1
= 1, b
2
= 0, b
3
= 0, b
n+4
= −
b
n
(n + 4)(n + 3)
.
The first few terms in the fundamental set of solutions are
w
1
= 1 −
1

12
z
4
+
1
672
z
8
− ··· , w
2
= z −
1
20
z
5
+
1
1440
z
9
− ··· .
In Figure 23.3 the five term approximation is graphed in a coarse dashed line, the ten term approximation is graphed
in a fine dashed line, and the numerical solution of w
1
is graphed in a solid line. The same is done for w
2
.
1196
1
2

3
4
5
6
-1
-0.5
0.5
1
1.5
1
2
3
4
5
6
-1
-0.5
0.5
1
1.5
Figure 23.3: The graph of approximations and numerical solution of w
1
and w
2
.
Result 23.1.1 Consider the n
th
order linear homogeneous equation
d
n

w
dz
n
+ p
n−1
(z)
d
n−1
w
dz
n−1
+ ··· + p
1
(z)
dw
dz
+ p
0
(z)w = 0.
If each of the coefficient functions p
i
(z) are analytic at z = z
0
then z
0
is an ordinary point
of the differential equation. The solution is analytic in some region containing z
0
and can
be expanded in a Taylor series. The radius of convergence of the series will be at least the

distance to the nearest singularity of the coefficient functions in the complex plane.
1197