3.5. Power Series Solutions

Power series solutions are another technique we can use to solve 2nd-order homogeneous ODEs of the form

(3.105)$$y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$$

This is useful for more-general cases where our other techniques fail.

For example, how would you find the solution to this ODE?

(3.106)$$(1+x^2)y^{\prime \prime }-4x{y}^{\prime }+6y=0$$

None of the methods we’ve discussed so far would allow us to find an analytical solution to this problem—but we can using a power series solution.

Power series solutions will be of the form

(3.107)$$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$$

where the coefficients $a_n$ are what we need to find.

1. First, for power series to be a valid solution, we need to check whether $x=0$ is an ordinary point of the ODE: is the ODE continuous and bounded at $x=0$?

Continuous means that there should be no discontinuity at $x=0$.

Bounded means that the solution should be finite at $x=0$.

For example, consider the ODE

(3.108)$$y^{\prime \prime }-4x{y}^{\prime }+(4x^2-2)y=0$$

Both $p(x)=-4x$ and $q(x)=(4x^2-2)$ are continuous and bounded at $x=0$, so $x=0$ is an ordinary point.

On the other hand, what about

(3.109)$$y^{\prime \prime }+x^3{y}^{\prime }+\frac{1}{x}y=0\text{ ?}$$

In this case, the solution is unbounded at $x=0$, and so it is not an ordinary point.

2. If $x=0$ is an ordinary point, then we can find a solution in the form of a power series:

(3.110)$$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$$

We then solve for the coefficients $a_n$ by plugging this in to the ODE. To do that, we’ll need to take advantage of certain properties of power series.

3.5.1. Properties of power series

• Dummy index rule. We can replace the index variable used in the power series with another index variable arbitrarily:

(3.111)$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\sum _{m=0}^{\mathrm{\infty }}{a}_m{x}^m$$

This is because the index variable is just a “dummy” that only has meaning inside the sum.

• Product rule. We can bring variables, including $x$, multiplying an entire power series into the power series:

(3.112)$$x\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+1}$$

• Derivatives. We can take derivatives of our power series:

(3.113)$$\begin{array}{rl}{y}^{\prime }& =\sum _{n=0}^{\mathrm{\infty }}{a}_n(n)x^{n-1}=\sum _{n=1}^{\mathrm{\infty }}{a}_n(n)x^{n-1}\\ y^{\prime \prime }& =\sum _{n=0}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}=\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}\end{array}$$

Notice that we can change where the sums in the power series start, because for $y^{\prime }$ the term corresponding to $n=1$ would just be zero, and similar for the first two terms of $y^{\prime \prime }$.

• Index shift. We can redefine the index used within a sum to shift where it starts. For example, if we let $m=n-1$, or $n=m+1$, then:

(3.114)$$\sum _{n=1}^{\mathrm{\infty }}{a}_n(n)x^{n-1}=\sum _{m=0}^{\mathrm{\infty }}{a}_{m+1}(m+1)x^m$$

Or, in other case, if we let $m=n-2$, or $n=m+2$, then:

(3.115)$$\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}=\sum _{m=0}^{\mathrm{\infty }}{a}_{m+2}(m+2)(m+1)x^m$$

Now, let’s apply these properties to solve ODEs.

3.5.2. Power series example 1

Let’s try to apply the power series approach to solve

(3.116)$$y^{\prime \prime }+y=0,$$

where we know the solution will be $y(x)=c_1\sin x+c_2\cos x$.

1. Is $x=0$ an ordinary point? Yes, the ODE is continuous and bounded at $x=0$. So, we can find a solution of the form $y(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$.

2. Now, we solve for the coefficents by plugging the power series into the ODE:

(3.117)$$\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=0$$

Let’s use the index shift rule on the first part of that:

(3.118)$$\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}\to \sum _{m=0}^{\mathrm{\infty }}{a}_{m+2}(m+2)(m+1)x^m$$

Then, we can use the dummy index rule to change $m$ back to $n$:

(3.119)$$\sum _{m=0}^{\mathrm{\infty }}{a}_m(m+2)(m+1)x^m\to \sum _{n=0}^{\mathrm{\infty }}{a}_n(n+2)(n+1)x^n$$

Now, let’s replace the first term in the ODE with that, merge both terms into a single sum, and simplify:

(3.120)$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{a}_n(n+2)(n+1)x^n+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n& =0\\ \sum _{n=0}^{\mathrm{\infty }}{x}^n[a_{n+2}(n+2)(n+1)+a_n]& =0\end{array}$$

There are infinite terms in this sum, involving the continuous variable $x$; the only way that equation can be satisfied is if

• $x=0$ always, which cannot be true, or

• $a_{n+2}(n+2)(n+1)+a_n=0$ for all values of $n$. This is what we can use to find the coefficients of our power series solution.

Use that expression to define a recursive formula for the coefficients:

(3.121)$$a_{n+2}=\frac{-a_n}{(n+1)(n+2)}$$

We can see that the even coefficients will be related to each other, and the odd coefficients will be related. Let’s try to identify a pattern with each, starting with the even terms:

(3.122)$$\begin{array}{rl}n=0:a_2& =\frac{-a_0}{1\cdot 2}=\frac{-a_0}{2!}\\ n=2:a_4& =\frac{-a_2}{3\cdot 4}=\frac{a_0}{4!}\\ n=4:a_6& =\frac{-a_4}{5\cdot 6}=\frac{-a_0}{6!}\end{array}$$

and the odd terms:

(3.123)$$\begin{array}{rl}n=1:a_3& =\frac{-a_1}{2\cdot 3}=\frac{-a_1}{3!}\\ n=3:a_5& =\frac{-a_3}{4\cdot 5}=\frac{a_1}{5!}\\ n=5:a_7& =\frac{-a_5}{6\cdot 7}=\frac{-a_1}{7!}\end{array}$$

Now, let’s put that all together:

(3.124)$$\begin{array}{rl}y(x)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\dots \\ y& =a_0(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\dots )+a_1(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots )\end{array}$$

which you might recognize as being the Taylor series expansion of sine and cosine:

(3.125)$$y(x)=a_0\cos x+a_1\sin x$$

So, our unknown coefficients end up being our integration constants, which we can use our two constraints to find.

3.5.3. Power series example 2

Find the solution to the ODE

(3.126)$$(1+x^2)y^{\prime \prime }-4x{y}^{\prime }+6y=0$$

First, rearrange into standard form:

(3.127)$$y^{\prime \prime }-\frac{4x}{1+x^2}{y}^{\prime }+\frac{6}{1+x^2}y=0$$

Then, check whether $x=0$ is an ordinary point: yes, it is.

Now, let’s insert the power series into the ODE:

(3.128)$$\begin{array}{rl}{y}^{\prime \prime }+x^2{y}^{\prime \prime }-4x{y}^{\prime }+6y& =0\\ \sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}+x^2\sum _{n=0}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}-4x\sum _{n=1}{a}_n(n)x^{n-1}+6\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n& =0\end{array}$$

First, we’ll use the power rule:

(3.129)$$\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}+\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^n-4\sum _{n=1}{a}_n(n)x^n+6\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=0$$

and then the index shift and dummy index rules on the first term:

(3.130)$$\sum _{n=2}^{\mathrm{\infty }}{a}_n(n)(n-1)x^{n-2}\to \sum _{m=0}^{\mathrm{\infty }}{a}_{m+2}(m+2)(m+1)x^m\to \sum _{n=0}^{\mathrm{\infty }}{a}_{n+2}(n+2)(n+1)x^n$$

Then, put that back into the full equation and combine the sums:

(3.131)$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{a}_{n+2}(n+2)(n+1)x^n+\sum _{n=0}^{\mathrm{\infty }}{a}_n(n)(n-1)x^n-4\sum _{n=1}{a}_n(n)x^n+6\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n& =0\\ \sum _{n=0}^{\mathrm{\infty }}{x}^n[a_{n+2}(n+2)(n+1)+a_n(n)(n-1)-4a_n(n)+6a_n]& =0\\ a_{n+2}(n+2)(n+1)+a_n(n^2-5n+6)& =0\\ a_{n+2}(n+2)(n+1)+a_n(n-3)(n-2)& =0\end{array}$$

Thus, our recursion formula for the coefficients $a_n$ is

(3.132)$$a_{n+2}=-a_n\frac{(n-3)(n-2)}{(n+1)(n+2)}$$

Again, we can see that the even terms will be related and the odd terms will be related:

(3.133)$$\begin{array}{rl}n=0:a_2& =-a_0\frac{6}{2}=-3a_0\\ n=2:a_4& =0\\ n=4:a_6& =-a_4\frac{2}{30}=0\\ & \dots \end{array}$$

and the odd terms:

(3.134)$$\begin{array}{rl}n=1:a_3& =-a_1\frac{2}{6}=\frac{-a_1}{3}\\ n=3:a_5& =0\\ n=5:a_7& =-a_5\frac{6}{42}=0\\ & \dots \end{array}$$

The solution is then

(3.135)$$\begin{array}{rl}y(x)& =a_0+a_{1}x-3a_0{x}^2-\frac{a_1}{3}{x}^3\\ y& =a_0(1-3x^2)+a_1(x-\frac{x^3}{3})\end{array}$$

where we find $a_0$ and $a_1$ using our initial or boundary conditions.