# 7.2: Series solutions of linear second order ODEs

### Introduction

Suppose we have a linear second order homogeneous ODE of the form

\[ p(x)y'' + q(x)y' + r(x)y = 0 \]

Suppose that \(p(x)\), \(q(x)\), and \(r(x)\) are polynomials. We will try a solution of the form

\[ y = \sum_{k=0}^\infty a_k\left(x-x_o\right)^2 \]

and solve for the \(a_k\) to try to obtain a solution defined in some interval around\(x_o\). The point \(x_o\) is called an **ordinary point **if \(p(x_o) \neq 0\). That is, the functions

\[ \dfrac{q(x)}{p(x)} \]

and

\[ \dfrac{r(x)}{p(x)} \]

are defined for \(x\) near \(x_o\). If \( p(x_0)=0\), then we say \(x_o\) is a singular point. Handling singular points is harder than ordinary points and so we now focus only on ordinary points.

Example 7.2.1: Let us start with a very simple example

\[ y'' - y = 0 \]

Let us try a power series solution near \(x_o=0\), which is an ordinary point. Every point is an ordinary point in fact, as the equation is constant coefficient. We already know we should obtain exponentials or the hyperbolic sine and cosine, but let us pretend we do not know this.

We try

\[ y = \sum_{k=0}^\infty a_k x^k \]

If we differentiate, the \( k=0 \) term is a constant and hence disappears. We therefore get

\[ y' = \sum_{k=1}^\infty k\, a_k \, x^{k-1} \]

We differentiate yet again to obtain (now the \(k=1\) term disappears)

\[ y'' = \sum_{k=2}^\infty k\, (k-1)\, a_k \, x^{k-2} \]

We reindex the series (replace \(k\) with \( k+2 \) ) to obtain

\[ y'' = \sum_{k=0}^\infty (k+2)\, (k+1)\, a_{k+2} \, x^k \]

Now we plug \(y\) and \(y''\) into the differential equation.

As is supposed to be equal to 0, we know that the coefficients of the resulting series must be equal to 0. Therefore,

The above equation is called a recurrence relation for the coefficients of the power series. It did not matter what or was. They can be arbitrary. But once we pick and , then all other coefficients are determined by the recurrence relation.

Let us see what the coefficients must be. First, and are arbitrary

So we note that for *even *\(k\), that is \( k=2n\) we get

\[ a_k = a_{2n} = \dfrac{a_o}{(2n)!}\]

and for *odd *\(k\) that is \( k=2n+1\) we have

\[ a_k = a_{2n+1} = \dfrac{a_1}{(2n+1)!}\]

Let us write down the series

We recognize the two series as the hyperbolic sine and cosine. Therefore,

\[ y= a_o \, \text{cosh} \, x + a_1\, \text{sinh}\, x \]

Of course, in general we will not be able to recognize the series that appears, since usually there will not be any elementary function that matches it. In that case we will be content with the series.

Example 7.2.2: Let us do a more complex example. Suppose we wish to solve **Airy’s equation**^{2}, that is

\[ y'' - xy = 0 \]

near the point \( x_o = 0 \), which is an ordinary point.

We try

\[ y = \sum_{k=0}^{\infty} a_k x^k \]

We differentiate twice (as above) to obtain

\[ y'' = \sum_{k=2}^{\infty} k(k-1)a_k x^{k-2} \]

We plug \(y\) into the equation

We reindex to make things easier to sum

Again \( y'' - xy\) is supposed to be 0 so first we notice that \(a_2 = 0 \) and also

Now we jump in steps of three. First we notice that since we must have that, , , , etc…. In general .

The constants and are arbitrary and we obtain

For where is a multiple of , that is we notice that

For where , we notice

In other words, if we write down the series for we notice that it has two parts

We define

and write the general solution to the equation as . Notice from the power series that and . Also, and . If we obtained a solution that satisfies the initial conditions and .

The functions and cannot be written in terms of the elementary functions that you know. See Figure 7.3 for the plot of the solutions and . These functions have many interesting properties. For example, they are oscillatory for negative (like solutions to ) and for positive they grow without bound (like solutions to ).

Sometimes a solution may turn out to be a polynomial.

Example 7.2.3: Let us find a solution to the so-called Hermite’s equation of order ^{3} is the equation

Let us find a solution around the point . We try

We differentiate (as above) to obtain

Now we plug into the equation

As we have

This recurrence relation actually includes (which comes about from ). Again and are arbitrary.

Let us separate the even and odd coefficients. We find that

Let us write down the two series, one with the even powers and one with the odd.

We then write

We also notice that if is a positive even integer, then is a polynomial as all the coefficients in the series beyond a certain degree are zero. If is a positive odd integer, then is a polynomial. For example, if , then

### 7.2.1 Exercises

In the following exercises, when asked to solve an equation using power series methods, you should find the first few terms of the series, and if possible find a general formula for the \(k^{th}\) coefficient.

Exercise 7.2.1: Use power series methods to solve \(y'' + y = 0 \) at the point .

Exercise 7.2.2: Use power series methods to solve \(y'' + 4xy = 0 \) at the point .

Exercise 7.2.3: Use power series methods to solve \(y'' - xy = 0 \) at the point .

Exercise 7.2.4: Use power series methods to solve \(y'' + x^2y = 0 \) at the point .

Exercise 7.2.5: The methods work for other orders than second order. Try the methods of this section to solve the first order system \(y' - xy = 0 \) at the point .

Exercise 7.2.6 (Chebyshev’s equation of order ): a) Solve using power series methods at . b) For what is there a polynomial solution?

Exercise 7.2.7: Find a polynomial solution to using power series methods.

Exercise 7.2.8: a) Use power series methods to solve at the point . b) Use the solution to part a) to find a solution for around the point .

Exercise 7.2.101: Use power series methods to solve at the point .

Exercise 7.2.102 (challenging): We can also use power series methods in nonhomogeneous equations. a) Use power series methods to solve at the point . Hint: Recall the geometric series. b) Now solve for the initial condition , .

Exercise 7.2.103: Attempt to solve at using the power series method of this section ( is a singular point). Can you find at least one solution? Can you find more than one solution?

^{2}Named after the English mathematician Sir George Biddell Airy (1801 – 1892).

^{3}Named after the French mathematician Charles Hermite (1822–1901).

### Contributors

- Jiří Lebl (Oklahoma State University).These pages were supported by NSF grants DMS-0900885 and DMS-1362337.