Power Series Solutions

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

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

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

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

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

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

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:

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.

Properties of power series¶

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

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:

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

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:

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

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

Power series example 1¶

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

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:

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

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

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

For this to be true for all $x$, each coefficient must be zero:

This recursion lets us generate all coefficients from $a_0$ and $a_1$ (the two initial conditions for a 2nd-order ODE).

Solution Construction¶

Let $a_0$ and $a_1$ be arbitrary constants:

• $a_2 = -\frac{a_0}{2\cdot 1}$

• $a_3 = -\frac{a_1}{3\cdot 2}$

• $a_4 = -\frac{a_2}{4\cdot 3} = \frac{a_0}{4\cdot 3\cdot 2\cdot 1}$

• $a_5 = -\frac{a_3}{5\cdot 4} = \frac{a_1}{5\cdot 4\cdot 3\cdot 2}$

So,

These are the Taylor series for $\cos x$ and $\sin x$! Thus, the power series solution recovers the known result.

Power series example 2¶

Find the solution to the ODE

First, rearrange into standard form:

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

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

First, we’ll use the power rule:

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

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

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

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

and the odd terms:

The solution is then

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