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)\[\begin{equation} y^{\prime\prime} + p(x) y^{\prime} + q(x) y = 0 \end{equation}\]

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)\[\begin{equation} (1 + x^2) y^{\prime\prime} - 4 x y^{\prime} + 6y = 0 \end{equation}\]

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)\[\begin{equation} y = \sum_{n=0}^{\infty} a_n x^n \end{equation}\]

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)\[\begin{equation} y^{\prime\prime} - 4xy^{\prime} + (4x^2 - 2)y = 0 \end{equation}\]

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)\[\begin{equation} y^{\prime\prime} + x^3 y^{\prime} + \frac{1}{x} y = 0 \text{ ?} \end{equation}\]

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)\[\begin{equation} y = \sum_{n=0}^{\infty} a_n x^n \end{equation}\]

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)\[\begin{equation} \sum_{n=0}^{\infty} a_n x^n = \sum_{m=0}^{\infty} a_m x^m \end{equation}\]

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)\[\begin{equation} x \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n x^{n+1} \end{equation}\]

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

(3.113)\[\begin{align} y^{\prime} &= \sum_{n=0}^{\infty} a_n (n) x^{n-1} = \sum_{n=1}^{\infty} a_n (n) x^{n-1} \\ y^{\prime\prime} &= \sum_{n=0}^{\infty} a_n (n)(n-1) x^{n-2} = \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} \end{align}\]

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)\[\begin{equation} \sum_{n=1}^{\infty} a_n (n) x^{n-1} = \sum_{m=0}^{\infty} a_{m+1} (m+1) x^m \end{equation}\]

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

(3.115)\[\begin{equation} \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} = \sum_{m=0}^{\infty} a_{m+2} (m+2)(m+1) x^m \end{equation}\]

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)\[\begin{equation} y^{\prime\prime} + y = 0 \;, \end{equation}\]

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 \(y(x) = \sum_{n=0}^{\infty} a_n x^n\).

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

(3.117)\[\begin{equation} \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} + \sum_{n=0}^{\infty} a_n x^n = 0 \end{equation}\]

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

(3.118)\[\begin{equation} \sum_{n=2}^{\infty} a_n (n)(n-1) x^{n-2} \rightarrow \sum_{m=0}^{\infty} a_{m+2} (m+2)(m+1) x^m \end{equation}\]

Then, we can use the dummy index rule to change \(m\) back to \(n\):

(3.119)\[\begin{equation} \sum_{m=0}^{\infty} a_m (m+2)(m+1) x^m \rightarrow \sum_{n=0}^{\infty} a_n (n+2)(n+1) x^n \end{equation}\]

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

(3.120)\[\begin{align} \sum_{n=0}^{\infty} a_n (n+2)(n+1) x^n + \sum_{n=0}^{\infty} a_n x^n &= 0 \\ \sum_{n=0}^{\infty} x^n \left[ a_{n+2}(n+2)(n+1) + a_n \right] &= 0 \end{align}\]

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)\[\begin{equation} a_{n+2} = \frac{-a_n}{(n+1)(n+2)} \end{equation}\]

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)\[\begin{align} n=0: \quad a_2 &= \frac{-a_0}{1 \cdot 2} = \frac{-a_0}{2!} \\ n=2: \quad a_4 &= \frac{-a_2}{3 \cdot 4} = \frac{a_0}{4!} \\ n=4: \quad a_6 &= \frac{-a_4}{5 \cdot 6} = \frac{-a_0}{6!} \end{align}\]

and the odd terms:

(3.123)\[\begin{align} n=1: \quad a_3 &= \frac{-a_1}{2 \cdot 3} = \frac{-a_1}{3!} \\ n=3: \quad a_5 &= \frac{-a_3}{4 \cdot 5} = \frac{a_1}{5!} \\ n=5: \quad a_7 &= \frac{-a_5}{6 \cdot 7} = \frac{-a_1}{7!} \end{align}\]

Now, let’s put that all together:

(3.124)\[\begin{align} y(x) &= a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \dots \\ y &= a_0 \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \right) + a_1 \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots \right) \end{align}\]

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

(3.125)\[\begin{equation} y(x) = a_0 \cos x + a_1 \sin x \end{equation}\]

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)\[\begin{equation} (1 + x^2) y^{\prime\prime} - 4x y^{\prime} + 6y = 0 \end{equation}\]

First, rearrange into standard form:

(3.127)\[\begin{equation} y^{\prime\prime} - \frac{4x}{1+x^2} y^{\prime} + \frac{6}{1+x^2} y = 0 \end{equation}\]

Then, check whether \(x=0\) is an ordinary point: yes, it is.

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

(3.128)\[\begin{align} y^{\prime\prime} + x^2 y^{\prime\prime} - 4 x y^{\prime} + 6 y &= 0 \\ \sum_{n=2}^{\infty} a_n (n)(n-1)x^{n-2} + x^2 \sum_{n=0}^{\infty} a_n (n)(n-1)x^{n-2} - 4 x \sum_{n=1} a_n (n) x^{n-1} + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \end{align}\]

First, we’ll use the power rule:

(3.129)\[\begin{equation} \sum_{n=2}^{\infty} a_n (n)(n-1)x^{n-2} + \sum_{n=2}^{\infty} a_n (n)(n-1)x^{n} - 4 \sum_{n=1} a_n (n) x^{n} + 6 \sum_{n=0}^{\infty} a_n x^n = 0 \end{equation}\]

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

(3.130)\[\begin{equation} \sum_{n=2}^{\infty} a_n (n)(n-1)x^{n-2} \rightarrow \sum_{m=0}^{\infty} a_{m+2} (m+2)(m+1) x^m \rightarrow \sum_{n=0}^{\infty} a_{n+2} (n+2)(n+1) x^n \end{equation}\]

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

(3.131)\[\begin{align} \sum_{n=0}^{\infty} a_{n+2} (n+2)(n+1) x^n + \sum_{n=0}^{\infty} a_n (n)(n-1)x^{n} - 4 \sum_{n=1} a_n (n) x^{n} + 6 \sum_{n=0}^{\infty} a_n x^n &= 0 \\ \sum_{n=0}^{\infty} x^n \left[ a_{n+2} (n+2)(n+1) + a_n (n)(n-1) - 4a_n (n) + 6a_n \right] &= 0 \\ a_{n+2} (n+2)(n+1) + a_n (n^2 -5n + 6) &= 0 \\ a_{n+2} (n+2)(n+1) + a_n (n-3)(n-2) &= 0 \\ \end{align}\]

Thus, our recursion formula for the coefficients \(a_n\) is

(3.132)\[\begin{equation} a_{n+2} = -a_n \frac{(n-3)(n-2)}{(n+1)(n+2)} \end{equation}\]

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

(3.133)\[\begin{align} n=0: \quad a_2 &= -a_0 \frac{6}{2} = -3 a_0 \\ n=2: \quad a_4 &= 0 \\ n=4: \quad a_6 &= -a_4 \frac{2}{30} = 0 \\ &\ldots \end{align}\]

and the odd terms:

(3.134)\[\begin{align} n=1: \quad a_3 &= -a_1 \frac{2}{6} = \frac{-a_1}{3} \\ n=3: \quad a_5 &= 0 \\ n=5: \quad a_7 &= -a_5 \frac{6}{42} = 0 \\ &\ldots \end{align}\]

The solution is then

(3.135)\[\begin{align} y(x) &= a_0 + a_1 x - 3 a_0 x^2 - \frac{a_1}{3} x^3 \\ y &= a_0 \left(1 - 3x^2 \right) + a_1 \left( x - \frac{x^3}{3} \right) \end{align}\]

where we find \(a_0\) and \(a_1\) using our initial or boundary conditions.