Chapter 5 Power Series Solution of ODEs

5 Introduction

Notes Proofread by Yunting Gao and corrections made on 04/05/2021

5.1.1 Review of Power Series

Convergence of Sequences and Series

I) We say a sequence of numbers {Sn }∞
n=1 converges to a number L ∈ R if lim Sn =
n→∞
L, i.e., the limit exists and the limit is the number L (which we call the limit of the
sequence).

A) If a sequence is increasing and bounded above, then it converges.

B) If a sequence is decreasing and bounded below, then it converges.

C) If Sk = f (k) where f is differentiable then lim Sn = lim f (x). This useful fact
n→∞ x→∞
allows us to find limits of some sequences using tools from calculus.
∞
X n
X
II) We say a series ak converges if the sequence of partial sums, Sn = ak con-
k=1 k=1
verges.

∞
a
converges to
 , if |r| < 1
1−r
X
A) A geometric series ark either
if |r| ≥ 1

diverges
k=0

B) If a series with nonnegative terms has all partial sums bounded, then the series
converges by the bounded monotone convergence theorem.
n
X
C) (Divergence test) If lim ak 6= 0 then the series ak diverges.
k→∞
k=1

D) (Integral Test) If ak = f (k) where f is continuous and eventually positive and

decreasing for x > a > 0, then

Z ∞ ∞
X
f (x) dx and ak either both converge or diverge.
a k=1

1
 
∞ converges , p>1

X 1
E) (p-test) The series =
kp 
k=0 diverges p≤1
F) (Direct Comparison) If 0 < dk ≤ ak ≤ ck , then
∞
X ∞
X
(i) If dk diverges ⇒ ak diverges.
k=1 k=1
∞
X ∞
X
(ii) If ck converges ⇒ ak converges.
k=1 k=1
ak
G) (Limit Comparison) If 0 < bk , ak and L = lim with 0 < L < ∞, then the series
k→∞ bk
∞
X ∞
X
ak and bk either both converge or both diverge.
k=1 k=1
 ∞
 X
L < 1 ⇒ series ak converges





 k=1


ak+1  ∞
H) (Ratio test) Let L = lim then L > 1 ⇒ series ak diverges .
X
k→∞ ak 


 k=1




L = 1 the test fails

p
k
I) (Root test) Let L = lim |ak | then we have the same result as above.
k→∞

J) (Alternating Series test) Assume {ak }∞k=1 are positive and (eventually) decreas-
X∞
ing to zero, then the alternating series (−1)k ak converges. For a convergent
k=1
alternating series we have |S − Sn | ≤ ak+1 .
∞
X ∞
X
K) If a series converges absolutely (i.e., |ak | converges) then ak converges.
k=1 k=1
The converse is not true as shown by the Harmonic series:

∞ ∞
X (−1)n X 1
converges but diverges.
n=1
n n=1
n

Taylor and MacLaurin Series

∞
X
An expression in the form an (x − a)n is called a Power Series centered at x = a.
n=0
An Analytic function f (x) is a function which possesses a convergent power series, i.e.,

2
 ∞
X
f (x) = an (x − a)n where the series converges absolutely on an interval |x − a| < R.
n=0
The number R is called the radius of convergence. If f is analytic then the coefficients an
must be given by the so called Taylor Coefficients and the series is called a Taylor series.
Namely we have
∞
X f (n) (a)
f (x) = (x − a)n .
n=0
n!

When a = 0 we call the Taylor series a MacLaurin Series. Such series tend to be a bit
simpler and so that is what we focus on for the most part in the following discussion.

Smooth Non-Analytic Functions

One might wonder whether there are smooth functions, i.e., functions for which all
derivatives of all order exist, which do not posses representation as a convergent power
series. The answer is yes! Here is an example.

e1/x2

if x 6= 0
f (x) = .
0 if x = 0


has all derivatives zero there. Consequently, the Taylor series of f(x) about x = 0 is iden-
tically zero. However, f(x) is not equal to the zero function, and so it is not equal to its
Taylor series around the origin.

Taylor Polynomial
Since in practice one cannot usually sum an infinite number of terms or find a closed
form for the sum it is often useful to consider truncating the infinite sum of a power series
which produces a polynomial. Sometimes this polynomial can be used as a accurate
approximation of the function.

Theorem 5.1 (Taylor’s Theorem with Remainder). Let N > 0 be an integer and f be a
function which is N times differentiable at a point x = a. Then we have

N
X f (n) (a)
f (x) = (x − a)n + RN (x)
n=0
n!

3
where Z x
1
RN (x) = (x − t)n f (N +1) (t) dt.
n! a

When a = 0 the Taylor series is called a MacLaurin Series.

Examples of Common MacLaurin Series

∞
X
f (x) an x n
n=0

∞
X xn
ex
n=0
n!
∞
X (−1)n x2n
cos(x)
n=0
(2n)!
∞
X (−1)n x2n+1
sin(x)
n=0
(2n + 1)!
∞
1 X
xn
1−x n=0

∞
X (−1)n xn+1
ln(1 + x)
n=0
(n + 1)
 
p p(p − 1) 2 p(p − 1)(p − 2) 3 p n
(1 + x) 1 + px + x + x + ··· + x
2! 3! n

Adding Power Series and Shifting Indices

∞
X
Given a power series f (x) = an xn , at least for x within the radius of convergence,
n=0
∞
X
0
we can differentiate the series to obtain f (x) = an nxn−1 . Furthermore we can repeat
n=0
this as many times as we like and we still obtain a convergent power series. So, for

4
 ∞
X
example, f 00 (x) = an n(n − 1)xn−2 . Suppose now we want to form an expression like
n=0

f 00 (x) + xf 0 (x) − f (x)

into a single power series, i.e., we want to add the series together. Then we would write

∞
X ∞
X ∞
X
an n(n − 1)xn−2 + x an nxn−1 − an x n .
n=0 n=0 n=0

This presents a problem since you cannot combine apples and oranges. By this we
mean that the powers of x in each sum are different and you cannot combine directly xn
with xn−2 for example. This problem can be easily fixed by Shifting Indices. Here is an
∞
X
example of what I mean. In the series an xn we can shift the index n inside the sum
n=0
down by two provided we shift the indices in the summation up by two to obtain exactly
the same sum, i.e.,
∞
X ∞
X
n
an x = an−2 xn−2 .
n=0 n=2

Similarly
∞
X ∞
X ∞
X
n−1 n
x an nx = an nx = an−2 (n − 2)xn−2 .
n=0 n=0 n=2

So we can write

∞
X ∞
X ∞
X
00 0 n−2 n−1
f (x) + xf (x) − f (x) = an n(n − 1)x +x an nx − an x n
n=0 n=0 n=0
∞
X ∞
X ∞
X
n−2 n−2
= an n(n − 1)x + an−2 (n − 2)x − an−2 xn−2
n=0 n=2 n=2
∞
X ∞
X
an n(n − 1)xn−2 + an−2 (n − 2) − 1 xn−2
 
=
n=0 n=2
X∞ X∞
= an n(n − 1)xn−2 + an−2 (n − 3)xn−2
n=0 n=2
∞
X ∞
X
−2 −1 n−2
= 0a0 x + 0a1 x + an n(n − 1)x + an−2 (n − 3)xn−2
n=2 n=2
∞
X
n(n − 1)an + (n − 3)an−2 xn−2
 
=
n=2

5
5.1.2 Power Series Solutions of ODEs

In this section we consider the problem of solving an ordinary differential equation of the
form
P (x)y 00 + Q(x)y 0 + R(x)y = 0 (1)
∞
X
in the form of a power series f (x) = an (x − a)n . In this discussion we will only consider
n=0
the simpler case in which a = 0 so we are looking for a solution as a MacLaurin Series.
So, in particular, we seek solution near x = 0.
In order that an equation in the form (1) have a solution which is analytic (has a con-
vergent power series) some assumptions must be made. First we must assume that P (x),
Q(x) and R(x) are analytic functions. In addition we must assume that P (x) is not zero at
or near x = 0. More general cases are considered in the next few sections of Chapter 5
but we will not have time to consider these cases.
Let us consider a very simple example that we solved back in Chapter 2 (a first order
linear example).

∞
X
0
Example 5.1. Find the general solution of y − y = 0 in the form y = an x n .
n=0
∞
X
First we compute y 0 = an nxn−1 and then we substitute these series into the differ-
n=0
ential equation and try to find coefficients an so that the resulting equation is satisfied. We
have

∞
X ∞
X
0 n−1
0=y −y = an nx − an x n
n=0 n=0
∞
X ∞
X
= an nxn−1 − an−1 xn−1
n=0 n=1
∞
X ∞
X
= 0a0 x−1 + an nxn−1 − an−1 xn−1
n=1 n=1
∞
X
−1
an n − an−1 xn−1 .
 
= 0a0 x +
n=1

The first term is zero for all nonzero values of x so we see that a0 can be any real

6
number, i.e. it is an arbitrary constant.
Now, in order that the remaining equation

∞
X
an n − an−1 xn−1 = 0
 
n=1

holds for all x 6= 0 we would need

an n − an−1 = 0, n = 1, 2, · · · .

This is the same as

an−1
an = , n = 1, 2, · · · (Recursion Formula).
n

The Recursion Formula can be used to successively obtain the terms an in terms of a0 as
follows.

a0 a1
n = 1, a1 = n = 2, a2 =
1 2

a0
=
2·1

a2 a3
n = 3, a3 = n = 4, a4 =
3 4

a0 a0
= =
3! 4!

It is easy to see the pattern and we can extrapolate the above to conclude that

a0
an = for all n = 1, 2, 3, · · · .
n!

So we have
∞ ∞ ∞
X
n
X a0 n
X xn
y = a0 + an x = x = a0 = a0 ex .
n=1 n=0
n! n=0
n!

From the first entry in our table of power series examples we see that y = a0 ex .

7
 Next we consider a slightly more complicated example which, once again, we could
easily solve using methods developed in Chapter 3.

∞
X
00
Example 5.2. Find the general solution of y + y = 0 in the form y = an x n .
n=0
∞
X ∞
X
First we compute y 0 = an nxn−1 and y 00 = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the
resulting equation is satisfied. We have

∞
X ∞
X
0 = y 00 + y = an n(n − 1)xn−2 + an x n
n=0 n=0
∞
X ∞
X
= an n(n − 1)xn−2 + an−2 xn−2
n=0 n=2
∞
X ∞
X
−2 −1 n−2
= 0a0 x + 0a1 x + an n(n − 1)x + an−2 xn−2
n=2 n=2
X∞
= 0a0 x−2 + 0a1 x−1 + an n(n − 1) + an−2 xn−2 .
 
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants.
Now, in order that the remaining equation

∞
X
an n(n − 1) + an−2 xn−2 = 0
 
n=1

holds for all x 6= 0 we would need

an n(n − 1) + an−2 = 0, n = 2, 3, · · · .

This is the same as

−an−2
an = , n = 2, 3, · · · (Recursion Formula).
n(n − 1)

The Recursion Formula can be used to successively obtain the terms an in terms of a0

8
and a1 as follows. In this present case we see that the

−a0 −a1
n = 2, a2 = n = 3, a3 =
2·1 3·2

−a0 −a1
= =
2! 3!

−a2 −a3
n = 4, a4 = n = 5, a5 =
4·3 5·4

(−1)2 a0 (−1)2 a1
= =
4! 5!

−a4 −a5
n = 6, a6 = n = 7, a7 =
6·5 7·6

(−1)3 a0 (−1)3 a1
= =
6! 7!

It is easy to see the two patterns that evolve, one for the even coefficients in terms of
a0 and the odd coefficients in terms of a1 . Every even integer can be written as n = 2k
for k = 0, 1, 2, · · · and every odd integer can be written as n = 2k + 1 for k = 1, 2, · · · .
Therefore we can write the even an terms for n ≥ 2 as

(−1)k a0
a2k = for all k = 1, 2, 3, · · · ,
(2k)!

and we can write the odd an terms for n ≥ 3 as

(−1)k a1
a2k+1 = for all k = 1, 2, 3, · · · .
(2k + 1)!

So we have
∞
X ∞
X ∞
X
y= an x n = a0 + a2k x2k + a1 x + a2k+1 x2k+1 .
n=0 k=1 k=1

So we have
∞ ∞
X (−1)k X (−1)k 2k+1
y = a0 + a0 x2k + a1 x + a1 x .
k=1
(2k)! k=1
(2k + 1)!

From the second and third entries in our table of power series examples we see that

9
y = a0 cos(x) + a1 sin(x).

Next lets consider an example with non-constant coefficients and an initial value prob-
lem. Once again this is an example that we could have solved back in Chapter 2 since it
is first order linear.

Example 5.3. Solve the initial value problem y 0 − 2xy = 0 with y(0) = 1 in the form of
∞
X
a series y = an x n .
n=0
∞
X
0
First we compute y = an nxn−1 and then we substitute these series into the differ-
n=0
ential equation and try to find coefficients an so that the resulting equation is satisfied. We
have

∞
X ∞
X
0 = y 0 − 2xy = an nxn−1 − 2 an xn+1
n=0 n=0
∞
X ∞
X
= an nxn−1 − 2an−2 xn−1
n=0 n=2
∞
X ∞
X
−1 0 n−1
= 0a0 x + a1 x + an nx − 2an−2 xn−1
n=2 n=2
∞
X
= 0a0 x−1 + a1 x0 + an n − 2an−2 xn−1 .
 
n=2

The first term is zero for all nonzero values of x so we see that a0 can be any real
number, i.e. it is an arbitrary constant. But the term a1 must be zero since the constant
on the left hand side of the equation is zero, i.e., a1 = 0.
Now, in order that the remaining equation

∞
X
an n − 2an−2 xn−1 = 0
 
n=2

holds for all x 6= 0 we would need

 
an n − 2an−2 = 0, n = 2, 3, · · · .

10
This is the same as

2an−2
an = , n = 2, 3, · · · (Recursion Formula).
n

The Recursion Formula can be used to successively obtain all the even terms an (for n
even) in terms of a0 and we can see that all the odd terms an (for n odd) must be 0 since
a1 = 0. Thus we know that
a2k+1 = 0, k = 1, 2, · · · .
∞
X
Note that if y = an xn then y(0) = a0 so we have a0 = 1 and
n=0

2a0
n = 2, a2 =
2

2a2 22 a0 22
n = 4, a4 = = =
4 4·2 4·2

2a4 23
n = 6, a6 = =
6 6·4·2

2a2k−2 2k
n = 2k, a2k = =
(2k) (2k) · (2k − 2) · · · 4 · 2

2k 1
= =
2k k! k!

It is easy to see the pattern and we can extrapolate the above to conclude that

1
a2k = for all k = 1, 2, · · · .
k!

We conclude that

∞ ∞ ∞
X
2k
X 1 2k X (x2 )k 2
y = a0 + a2k x = a0 + a0 x = = ex .
k=1 k=1
k! k=0
k!

Where we have used the first entry in our table of power series examples with x2 instead
of x.
Notice this is a first order linear initial value problem. We could find a general solution

11
 2
using the integration factor e−x . to get

h 2
i0 2 2
e−x y = 0 ⇒ e−x y = C ⇒ y = Cex .

Example 5.4. Find the general solution of (x2 + 1)y 00 − 4xy 0 + 6y = 0 in the form
∞
X
y= an xn . Notice that near x = 0 the leading coefficient is not zero and all coefficients
n=0
are analytic so we can expect to have a power series solution.
X∞ X∞
0 n−1 00
First we compute y = an nx and y = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the
resulting equation is satisfied. We have

∞
X ∞
X ∞
X
2 00 0 2 n−2 n−1
0 = (x + 1)y − 4xy + 6y = (x + 1) an n(n − 1)x − 4x an nx +6 an x n
n=0 n=0 n=0
∞
X ∞
X
an n(n − 1)xn−2 + n(n − 1) − 4n + 6 an xn
 
=
n=0 n=0
∞
X ∞
X
−2 −1 n−2
= 0a0 x + 0a1 x + an n(n − 1)x + (n2 − 5n + 6)an xn
n=2 n=0
∞
X ∞
X
= 0a0 x−2 + 0a1 x−1 + an n(n − 1)xn−2 + (n2 − 9n + 20)an−2 xn−2
n=2 n=2
X∞
= 0a0 x−2 + 0a1 x−1 + an n(n − 1) + (n − 4)(n − 5)an−2 xn−2 .
 
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants.
Now, in order that the remaining equation

∞
X
an n(n − 1) + (n − 4)(n − 5)an−2 xn−2 = 0
 
n=2

holds for all x 6= 0 we would need

an n(n − 1) + (n − 4)(n − 5)an−2 = 0, n = 2, 3, · · · .

12
This is the same as

−(n − 4)(n − 5)an−2

an = , n = 2, 3, · · · (Recursion Formula).
n(n − 1)

The Recursion Formula can be used to successively obtain the terms an in terms of a0
and a1 as follows. In this present case we see that the

−(−2)(−3)a0 −(−1)(−2)a1
n = 2, a2 = n = 3, a3 =
2·1 3·2

−a1
= −3a0 =
3

−(0)(−1)a2 −(1)(0)a3
n = 4, a4 = n = 5, a5 =
4·3 5·4

=0 =0

n = 6, a6 = 0 n = 7, a7 = 0

.. .. .. ..
. . . .

It is easy to see that for n = 2k with k ≥ 2 we have a2k = 0 and for n = 2k + 1 with
k ≥ 2 we have a2k+1 = 0. So we have

∞  
X
n 2 1 3
y= an x = a0 (1 − 3x ) + a1 x− x .
n=0
3

In the next example we consider a problem which looks very much like the previous
one but the answer ends up quite different.

Example 5.5. Find the general solution of (1 − x2 )y 00 − 6xy 0 − 4y = 0 in the form

∞
X
y= an xn . Notice that near x = 0 the leading coefficient is not zero and all coefficients
n=0
are analytic so we can expect to have a power series solution.

13
 ∞
X ∞
X
First we compute y 0 = an nxn−1 and y 00 = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the
resulting equation is satisfied. We have

0 = (1 − x2 )y 00 − 6xy 0 − 4y
∞
X ∞
X ∞
X
= (1 − x2 ) an n(n − 1)xn−2 − 6x an nxn−1 − 4 an x n
n=0 n=0 n=0
∞
X ∞
X
n−2
n(n − 1) + 6n + 4 an xn
 
= an n(n − 1)x −
n=0 n=0
∞
X ∞
X
−2 −1 n−2
= 0a0 x + 0a1 x + an n(n − 1)x − (n2 + 5n + 4)an xn
n=2 n=0
∞
X ∞
X
= 0a0 x−2 + 0a1 x−1 + an n(n − 1)xn−2 − (n2 + n − 2)an−2 xn−2
n=2 n=2
∞
X ∞
X
= 0a0 x−2 + 0a1 x−1 + an n(n − 1)xn−2 − (n − 1)(n + 2)an−2 xn−2
n=2 n=2
X∞
= 0a0 x−2 + 0a1 x−1 + (n − 1) an n + (n + 2)an−2 xn−2 .
 
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants.
Now, in order that the remaining equation

∞
X
(n − 1) an n − (n + 2)an−2 xn−2 = 0
 
n=2

holds for all x 6= 0 we would need

an n − (n + 2)an−2 = 0, n = 2, 3, · · · .

This is the same as

(n + 2)an−2
an = , n = 2, 3, · · · (Recursion Formula).
n

The Recursion Formula can be used to successively obtain the terms an in terms of a0

14
and a1 as follows. In this present case we see that the

4 a0 5 a1
n = 2, a2 = n = 3, a3 =
2 3

6a2 7 a1
n = 4, a4 = n = 5, a5 =
4 5

6 · 4 a0 7 · 5 a1
= =
4·2 5·3

8 a4 9 a5
n = 6, a6 = n = 7, a7 =
6 7

8 · 6 · 4 a0 9 · 7 · 5 a1
= =
6·4·2 7·5·3

It is easy to see that for n = 2k with k ≥ 2 we have

(2k + 2)(2k) · · · (6)(4) 2k (k + 1)!

a2k = a0 = a0 = (k + 1) a0
(2k)(2k − 2) · · · (4)(2) 2k k!

and for n = 2k + 1 with k ≥ 2 we have

(2k + 3)(2k + 1) · · · (7)(5) (2k + 3)

a2k+1 = a1 = a1 .
(2k + 1)(2k − 1) · · · (5)(3) 3

So we have

∞
" ∞
# " ∞
#
X
n
X
2k 1X 2k+1
y= an x = a0 + (k + 1) x + a1 x + (2k + 3) x .
n=0 k=1
3 k=1

With a little work it can be shown (using geometric series arguments) that for |x| < 1 the
above sums reduce to
1 (3x − x3 )
y = a0 + a 1 .
(1 − x2 )2 3(1 − x2 )2
Namely, let us define v = x2 so that

∞
X ∞
X
y 1 = a0 + a0 (k + 1)x2k = (k + 1)v k
k=1 k=0

15
 ∞
!  
d X
k+1 d v
= a0 v =
dv k=0
dv 1−v
1 · (1 − v) − v · (−1) 1
= a0 2
=
(1 − v) (1 − v)2
a0
= .
(1 − x2 )2

For y2 we have

∞
1X
y2 = a1 x + a1 (2k + 3)x2k+1
3 k=1
∞ ∞
1X 1 X
= a1 (2k + 3)x2k+1 = a1 (2k + 3)x2k+2
3 k=0 3x k=0
∞
! ∞
!
1 d X 2k+3 1 d X
= a1 x = a1 x3 (x2 )k
3x dx k=0 3x dx k=0
x3 1 3x2 − x4
   
1 d
= a1 = a1
3x dx 1 − x2 3x (1 − x2 )2
3x − x3
= a1 .
3(1 − x2 )2

Next we consider a second order IVP.

Example 5.6. Find the solution of the IVP y 00 − 2y 0 + y = 0 with y(0) = 0 and y 0 (0) = 1
∞
X
in the form y = an xn . Notice that near x = 0 the leading coefficient is not zero and all
n=0
coefficients are analytic so we can expect to have a power series solution.
X∞ ∞
X
First we compute y 0 = an nxn−1 and y 00 = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the
resulting equation is satisfied. We have

0 = y 00 − 2y 0 + y
∞
X ∞
X ∞
X
= an n(n − 1)xn−2 − 2 an nxn−1 + an x n
n=0 n=0 n=0
∞
X ∞
X ∞
X
= an n(n − 1)xn−2 − 2 an nxn−1 + an x n
n=0 n=0 n=0
∞
X ∞
X ∞
X
−2 −1 n−2 n−2
= 0a0 x + 0a1 x + an n(n − 1)x + an−1 (n − 1)x + an−2 xn−2
n=2 n=1 n=2

16
 ∞
X ∞
X ∞
X
= 0a0 x−2 + 0a1 x−1 + + an n(n − 1)xn−2 − 2a0 (0)x−1 + −2 an−1 (n − 1)xn−2 + an−2 xn−2
n=2 n=2 n=2
∞
X
−2 −1
= 0a0 x + [0a1 − 2a0 (0)]x + [an n(n − 1) − 2an−1 (n − 1) + an−2 ]xn−2
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants. But according to the given initial
conditions we can say that a0 = 0 and a1 = 1.
Now, in order that the remaining equation

∞
X
[an n(n − 1) − 2an−1 (n − 1) + an−2 ]xn−2 = 0
n=2

holds for all x 6= 0 we would need

an n(n − 1) − 2an−1 (n − 1) + an−2 = 0, n = 2, 3, · · · .

This is the same as

2(n − 1)an−1 − an−2

an = , n = 2, 3, · · · (Recursion Formula).
n(n − 1)

The Recursion Formula can be used to successively obtain the terms an in terms of a0
and a1 as follows. In this present case we see that the coefficients do no break up into the
even and odd terms.

2(1)a1 − a0
n = 2, a2 = =1
2·1
2(2)a2 − a1 4−1 1
n = 3, a3 = = =
3·2 3·2 2!
2(3)(1/2!) − 1 1
n = 4, a4 = =
4·3 3!
2(4)a4 − a3 5 1
n = 5, a4 = = =
5·4 5! 4!

17
 Writing out a few more if necessary you can conclude that

1
an = n ≥ 2.
(n − 1)!

Thus we obtain

∞ ∞
X
n xn X
y = a0 + a1 x + an x = x +
n=2 n=2
(n − 1)!
!
X xn+1 X xn
=x+ =x 1+
n=1
n! n=1
n!
X xn
=x = xex .
n=0
n!

Notice back in Chapter 3 when we worked a problem like this it went as follows. Con-
sider the auxiliary equation
0 = r2 − 2r + 1 = (r − 1)2

which implies r = 1, 1, a double root. Therefore y = c1 ex + c2 xex and we need y 0 =

c1 ex + c2 (x + 1)ex . Apply the initial conditions to get c1 = 0 and c2 = 1 so we have y = xex .
∞
X
Example 5.7. Find the general solution of 00 0
y − xy − y = 0 in the form y = an x n .
n=0
Notice that near x = 0 the leading coefficient is not zero and all coefficients are analytic
so we can expect to have a power series solution.
X∞ X ∞
0 n−1 00
First we compute y = an nx and y = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the
resulting equation is satisfied. We have

0 = y 00 − xy 0 − y
∞
X ∞
X ∞
X
n−2 n−1
= an n(n − 1)x −x an nx − an x n
n=0 n=0 n=0
∞
X ∞
X ∞
X
= an n(n − 1)xn−2 − an nxn + an x n
n=0 n=0 n=0

18
 ∞
X ∞
X
= an n(n − 1)xn−2 − (n + 1)an xn
n=0 n=0
∞
X ∞
X
−2 −1 n−2
= 0a0 x + 0a1 x + an−1 (n − 1)x − (n − 1)an−2 xn−2
n=0 n=2
X∞
= 0a0 x−2 + 0a1 x−1 + [an n(n − 1) − (n − 1)an−2 ]xn−2
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants. But according to the given initial
conditions we can say that a0 = 0 and a1 = 1.
Now, in order that the remaining equation

∞
X
[an n(n − 1) − (n − 1)an−2 ]xn−2 = 0
n=2

holds for all x 6= 0 we would need

 
an n(n − 1) − (n − 1)an−2 = 0, n = 2, 3, · · · .

This is the same as

an−2
an = , n = 2, 3, · · · (Recursion Formula).
n

The Recursion Formula can be used to successively obtain the terms an in terms of a0
and a1 as follows. In this present case we see that the
Once again as in some earlier examples the coefficients break up into the even and
odd terms.

a0
n = 2, a2 =
2
a2 a0
n = 4, a4 = =
4 4·2
a4 a0
n = 6, a6 = =
6 6·4·2

19
 a2k−2 a0
n = 2k, a2k = = k
2k 2 k!

Writing out a few more if necessary you can conclude that

a0
a2k = k ≥ 1.
2k k!

a1
n = 3, a3 =
3
a3 a1
n = 5, a5 = =
5 5·3
a5 a1
n = 7, a7 = =
7 7·5·1
a2k−1 a1
n = 2k + 1, a2k+1 = =
2k + 1 (2k + 1)(2k − 1) · · · 5 · 3
2k k!a1
=
(2k + 1)!

The solution can be written in the form

y = a0 y1 + a1 y2

where  2 k
∞ ∞ x
X a0 x2k X 2 2 /2
y 1 = a0 + = a0 = ex .
k=1
2k k! k=0
k!

The odd terms present a bigger challenge and can only give a function in closed form
in terms of the error function
Z x
2 2
erf(x) = √ e−t dt.
π 0

We get  
x 2 /2
y2 = erf √ ex .
2
Let us consider an example that could easily be on an exam.

20
Example 5.8. Find the solution of the IVP (2 − x)y 0 + 3y = 0 with y(0) = 8 in the form
∞
X
y = an xn . All the coefficients are analytic but the leading coefficient can be 0 when
n=0
x = 2 so we need to make sure we say away from x = 2 but otherwise we can expect to
have a convergent power series solution in −2 < x < 2.
∞
X
0
First we compute y = an nxn−1 . Then we substitute these series into the differential
n=0
equation and try to find the coefficients an so that the resulting equation is satisfied. We
have

∞
X ∞
X
0 = (2 − x) an nxn−1 + 3 an x n
n=0 n=0
∞
X ∞
X ∞
X
n−1 n
=2 an nx − an nx + 3 an x n
n=0 n=0 n=0
∞
X ∞
X
= 2an nxn−1 − an (n − 3)xn
n=0 n=0
∞
X ∞
X
= 2an nxn−1 − an−1 (n − 4)xn−1
n=0 n=1
∞
X
= 2a0 · 0x−1 + [2an n − (n − 4)an−1 ]xn−1
n=1

From this we conclude a0 is arbitrary and [2an n − (n − 4)an−1 ] = 0 for all n ≥ 1. Thus we
obtain the recursion formula

(n − 4)an−1
an = n = 1, 2, · · · .
2n

Together with the initial condition a0 = y(0) = 8 this gives us

a0 (1 − 4)
n = 1, a1 = = −12
2·1
a1 (2 − 4) −12(−2)
n = 2, a2 = = =6
2·2 2·2
a2 (3 − 4) −6
n = 3, a3 = = = −1
2·3 6
a3 (4 − 4)
n = 4, a4 = =0
2·4

21
 Therefore all an = 0 for n ≥ 4. Our series solution becomes a polynomial

y = a0 + a1 x + a2 x2 + a3 x3 = 8 − 12x + 6x2 − x3 = (2 − x)3 .

Here is a another very similar looking example.

Example 5.9. Find the solution of the IVP (2 + x)y 0 + y = 0 with y(0) = 1 in the form
∞
X
y = an xn . All the coefficients are analytic but the leading coefficient can be 0 when
n=0
x = −2 so we need to make sure we say away from x = 2 but otherwise we can expect to
have a convergent power series solution in −2 < x < 2.
∞
X
0
First we compute y = an nxn−1 . Then we substitute these series into the differential
n=0
equation and try to find the coefficients an so that the resulting equation is satisfied. We
have

∞
X ∞
X
n−1
0 = (2 + x) an nx + an x n
n=0 n=0
∞
X ∞
X ∞
X
=2 an nxn−1 + an nxn + an x n
n=0 n=0 n=0
∞
X ∞
X
= 2an nxn−1 + an (n + 1)xn
n=0 n=0
X∞ X∞
= 2an nxn−1 + an−1 (n − 4)xn−1
n=0 n=1
∞
X
= 2a0 · 0x−1 + [2an n − nan−1 ]xn−1
n=1

From this we conclude a0 is arbitrary and [2an n + nan−1 ] = 0 for all n ≥ 1. Thus we obtain
the recursion formula
−an−1
an = n = 1, 2, · · · .
2
Together with the initial condition a0 = y(0) = 1 this gives us

22
 −a0 1
n = 1, a1 = =−
2 2
 2
−a1 −1
n = 2, a2 = =
2 2
 3
−a2 −1
n = 3, a3 = =
2 2
 4
−a3 ) −1
n = 4, a4 = =
2 2

So in general we get  n
−1
an = n = 1, 2, · · · .
2
Therefore all an = 0 for n ≥ 4. Our series solution becomes a polynomial

∞ ∞  n ∞  n
X
n
X −1 n
X −x
y= an x = 1 + x = .
n=0 n=1
2 n=0
2

 
−x
This is a geometric series with common ration which is convergent so long as
2
−x
< 1 or |x| < 2. And we get
2

1 2
y= −x

= .
1− 2
2+x

We consider one final example from mathematical physics - the Airy Equation. Like the
last example, this is a problem that we could not have solved using any previous methods
covered in the class. As simple as it looks this is example produces the most complicated
answer we have considered thus far.
∞
X
00
Example 5.10. Find the general solution of y − xy = 0 in the form y = an xn . All
n=0
the coefficients are analytic and we can expect to have a power series solution.
X∞ X∞
0 n−1 00
First we compute y = an nx and y = an n(n − 1)xn−2 . Then we substitute
n=0 n=0
these series into the differential equation and try to find the coefficients an so that the

23
resulting equation is satisfied. We have

∞
X ∞
X
00 n−2
0 = y − xy = an n(n − 1)x −x an x n
n=0 n=0
∞
X ∞
X
= an n(n − 1)xn−2 − an xn+1
n=0 n=0
∞
X ∞
X
−2 −1 0 n−2
= 0a0 x + 0a1 x + 2a2 x + an n(n − 1)x − an−3 xn−2
n=3 n=3
X∞
= 0a0 x−2 + 0a1 x−1 + 2a2 x0 + an n(n − 1) − an−3 xn−2 .
 
n=2

The first two terms are zero for all nonzero values of x so we see that a0 and a1 can
be any real numbers, i.e. they are arbitrary constants. But, in order that the third term be
zero we must have a2 = 0.
Then, in order that the remaining equation

∞
X
an n(n − 1) − an−3 xn−2 = 0
 
n=2

holds for all x 6= 0 we would need

 
an n(n − 1) − an−3 = 0, n = 2, 3, · · · .

This is the same as

an−3
an = , n = 3, 4, · · · (Recursion Formula).
n(n − 1)

The Recursion Formula can be used to successively obtain the terms an in terms of a0
and a1 as follows. In this present case we see that the a0 and a1 are arbitrary constants
and a2 = 0.
For this example we see that the coefficients break up into three groups:

1. terms that begin with a0 and differ by an index of 3, i.e., a3k for k = 1, 2, · · · .

2. terms that begin with a1 and differ by an index of 3, i.e., a3k+1 for k = 1, 2, · · · .

24
 3. terms that begin with a2 and differ by an index of 3, i.e., a3k+2 for k = 1, 2, · · · .

Notice that since a2 = 0 we see that all the terms in the group a3k+2 = 0. So we only
need to consider terms of the form a3k and terms of the form a3k+1 for k = 0, 1, · · · .
With some work we can compute the general terms in each case:

1. For k = 1, 2, ·
a0
a3k = .
(2 · 3)(5 · 6) · · · (3k − 1)(3k)

2. For k = 1, 2, ·
a1
a3k+1 = .
(3 · 4)(6 · 7) · · · (3k)(3k + 1)

So we have
y = a0 y1 + a1 y2

where " ∞
#
3k
X x
y1 = 1 +
k=1
(2 · 3)(5 · 6) · · · (3k − 1)(3k)

and " ∞
#
X x3k+1
y2 = x + .
k=1
(3 · 4)(6 · 7) · · · (3k)(3k + 1)

y1 y2

25