
# 7.2: Series solutions of linear second order ODEs

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

$y = \sum_{k=0}^\infty a_k {(x-x_0)}^k$

and solve for the $$a_k$$ to try to obtain a solution defined in some interval around $$x_0$$. The point $$x_0$$ is called an \emph{\myindex{ordinary point}} if $$p(x_0) \not= 0$$. That is, the functions

$\frac{q(x)}{p(x)} \qquad \text{and} \qquad \frac{r(x)}{p(x)}$

are defined for $$x$$ near $$x_0$$. If $$p(x_0) = 0$$, then we say $$x_0$$ is a singular point. Handling singular points is harder than ordinary points and so we now focus only on ordinary points. \begin{example} Let us start with a very simple example

$y'' - y = 0 .$

Let us try a power series solution near $$x_0 = 0$$, 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

$\begin{split} 0 = y''-y & = \Biggl( \sum_{k=0}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ & = \sum_{k=0}^\infty \,\Bigl( (k+2)\,(k+1) \, a_{k+2} x^k - a_k x^k \Bigr) \\ & = \sum_{k=0}^\infty \,\bigl( (k+2)\,(k+1) \,a_{k+2} - a_k \bigr) \, x^k . \end{split}$

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

$(k+2)\,(k+1) \,a_{k+2} - a_k = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{a_k}{(k+2)(k+1)} .$

The above equation is called a \emph{\myindex{recurrence relation}} for the coefficients of the power series. It did not matter what $$a_0$$ or $$a_1$$ was. They can be arbitrary. But once we pick $$a_0$$ and $$a_1$$, then all other coefficients are determined by the recurrence relation. Let us see what the coefficients must be. First, $$a_0$$ and $$a_1$$ are arbitrary

$a_2 = \frac{a_0}{2}, \quad a_3 = \frac{a_1}{(3)(2)}, \quad a_4 = \frac{a_2}{(4)(3)} = \frac{a_0}{(4)(3)(2)}, \quad a_5 = \frac{a_3}{(5)(4)} = \frac{a_1}{(5)(4)(3)(2)}, \quad \ldots$

So we note that for even $$k$$, that is $$k=2n$$ we get

$a_k = a_{2n} = \frac{a_0}{(2n)!} ,$ and for odd $$k$$, that is $$k=2n+1$$ we have $a_k = a_{2n+1} = \frac{a_1}{(2n+1)!} .$

Let us write down the series

$y = \sum_{k=0}^\infty a_k x^k = \sum_{n=0}^\infty \left( \frac{a_0}{(2n)!} \,x^{2n} + \frac{a_1}{(2n+1)!} \,x^{2n+1} \right) = a_0 \sum_{n=0}^\infty \frac{1}{(2n)!} \,x^{2n} + a_1 \sum_{n=0}^\infty \frac{1}{(2n+1)!} \,x^{2n+1} .$

We recognize the two series as the hyperbolic sine and cosine. Therefore, $y = a_0 \cosh x + a_1 \sinh x .$ \end{example} 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. \begin{example} Let us do a more complex example. Suppose we wish to solve \emph{\myindex{Airy's equation}}% \footnote{Named after the English mathematician Sir George Biddell Airy (1801--1892), that is

$y'' - xy = 0 ,$

near the point $$x_0 = 0$$. Note that $$x_0 = 0$$ 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 $\begin{split} 0 = y''-xy &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - x \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - \Biggl( \sum_{k=0}^\infty a_k x^{k+1} \Biggr) . \end{split}$ We reindex to make things easier to sum $\begin{split} 0 = y''-xy &= \Biggl( 2 a_2 + \sum_{k=1}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=1}^\infty a_{k-1} x^k \Biggr) . \\ &= 2 a_2 + \sum_{k=1}^\infty \Bigl( (k+2)\,(k+1) \, a_{k+2} - a_{k-1} \Bigr) \, x^k . \end{split}$ Again $$y''-xy$$ is supposed to be 0 so first we notice that $$a_2 = 0$$ and also $(k+2)\,(k+1) \,a_{k+2} - a_{k-1} = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{a_{k-1}}{(k+2)(k+1)} .$ Now we jump in steps of three. First we notice that since $$a_2 = 0$$ we must have that, $$a_5 = 0$$, $$a_8 = 0$$, $$a_{11}=0$$, etc\ldots. In general $$a_{3n+2} = 0$$. The constants $$a_0$$ and $$a_1$$ are arbitrary and we obtain $a_3 = \frac{a_0}{(3)(2)}, \quad a_4 = \frac{a_1}{(4)(3)}, \quad a_6 = \frac{a_3}{(6)(5)} = \frac{a_0}{(6)(5)(3)(2)}, \quad a_7 = \frac{a_4}{(7)(6)} = \frac{a_1}{(7)(6)(4)(3)}, \quad \ldots$ For $$a_k$$ where $$k$$ is a multiple of $$3$$, that is $$k=3n$$ we notice that $a_{3n} = \frac{a_0}{(2)(3)(5)(6) \cdots (3n-1)(3n)} .$ For $$a_k$$ where $$k = 3n+1$$, we notice $a_{3n+1} = \frac{a_1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} .$ In other words, if we write down the series for $$y$$ we notice that it has two parts $\begin{split} y &= \left( a_0 + \frac{a_0}{6} x^3 + \frac{a_0}{180} x^6 + \cdots + \frac{a_0}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots \right) \\ &\phantom{=} + \left( a_1 x + \frac{a_1}{12} x^4 + \frac{a_1}{504} x^7 + \cdots + \frac{a_1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots \right) \\ & = a_0 \left( 1 + \frac{1}{6} x^3 + \frac{1}{180} x^6 + \cdots + \frac{1}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots \right) \\ &\phantom{=} + a_1 \left( x + \frac{1}{12} x^4 + \frac{1}{504} x^7 + \cdots + \frac{1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots \right) . \end{split}$

We define

\begin{align*} y_1(x) &= 1 + \frac{1}{6} x^3 + \frac{1}{180} x^6 + \cdots + \frac{1}{(2)(3)(5)(6) \cdots (3n-1)(3n)} x^{3n} + \cdots, \\ y_2(x) &= x + \frac{1}{12} x^4 + \frac{1}{504} x^7 + \cdots + \frac{1}{(3)(4)(6)(7) \cdots (3n)(3n+1)} x^{3n+1} + \cdots , \end{align*}

and write the general solution to the equation as $$y(x)= a_0 y_1(x) + a_1 y_2(x)$$. Notice from the power series that $$y_1(0) = 1$$ and $$y_2(0) = 0$$. Also, $$y_1'(0) = 0$$ and $$y_2'(0) = 1$$. Therefore $$y(x)$$ is a solution that satisfies the initial conditions $$y(0) = a_0$$ and $$y'(0) = a_1$$. \begin{figure}[h!t] \capstart \begin{center} \diffyincludegraphics{width=3in}{width=4.5in}{ps-airy} \caption{The two solutions $$y_1$$ and $$y_2$$ to Airy's equation.\label{ps:airyfig}} \end{center} \end{figure} \end{example} The functions $$y_1$$ and $$y_2$$ cannot be written in terms of the elementary functions that you know. See \figureref{ps:airyfig} for the plot of the solutions $$y_1$$ and $$y_2$$. These functions have many interesting properties. For example, they are oscillatory for negative $$x$$ (like solutions to $$y''+y=0) and for positive \(x$$ they grow without bound (like solutions to $$y''-y=0). \medskip Sometimes a solution may turn out to be a polynomial. \begin{example} Let us find a solution to the so-called \emph{\myindex{Hermite's equation of order \(n}}% \footnote{Named after the French mathematician Charles Hermite} (1822--1901) is the equation $y'' -2xy' + 2n y = 0 .$ Let us find a solution around the point \(x_0 = 0$$. We try

$y = \sum_{k=0}^\infty a_k x^k .$

We differentiate (as above) to obtain

\begin{align*} y' &= \sum_{k=1}^\infty k a_k x^{k-1} , \\ y'' &= \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} . \end{align*}

Now we plug into the equation

$\begin{split} 0 = y''-2xy'+2ny &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - 2x \Biggl( \sum_{k=1}^\infty k a_k x^{k-1} \Biggr) + 2n \Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) \\ &= \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) - \Biggl( \sum_{k=1}^\infty 2k a_k x^k \Biggr) + \Biggl( \sum_{k=0}^\infty 2n a_k x^k \Biggr) \\ &= \Biggl(2a_2+ \sum_{k=1}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) - \Biggl( \sum_{k=1}^\infty 2k a_k x^k \Biggr) + \Biggl( 2na_0 + \sum_{k=1}^\infty 2n a_k x^k \Biggr) \\ &= 2a_2+2na_0+ \sum_{k=1}^\infty \bigl( (k+2)\,(k+1) \, a_{k+2} - 2ka_k + 2n a_k \bigr) x^k . \end{split}$

As $$y''-2xy'+2ny = 0$$

we have

$(k+2)\,(k+1) \, a_{k+2} + ( - 2k+ 2n) a_k = 0 , \qquad \text{or} \qquad a_{k+2} = \frac{(2k-2n)}{(k+2)(k+1)} a_k .$

This recurrence relation actually includes $$a_2 = -na_0$$ (which comes about from $$2a_2+2na_0 = 0). Again \(a_0$$ and $$a_1$$ are arbitrary.

\begin{align*} & a_2 = \frac{-2n}{(2)(1)}a_0, \qquad a_3 = \frac{2(1-n)}{(3)(2)} a_1, \\ & a_4 = \frac{2(2-n)}{(4)(3)} a_2 = \frac{2^2(2-n)(-n)}{(4)(3)(2)(1)} a_0 , \\ & a_5 = \frac{2(3-n)}{(5)(4)} a_3 = \frac{2^2(3-n)(1-n)}{(5)(4)(3)(2)} a_1 , \quad \ldots \end{align*}

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

\begin{align*} a_{2m} &=\frac{2^m(-n)(2-n)\cdots(2m-2-n)}{(2m)!} , \\ a_{2m+1} &=\frac{2^m(1-n)(3-n)\cdots(2m-1-n)}{(2m+1)!} . \end{align*}

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

\begin{align*} y_1(x) & = 1+\frac{2(-n)}{2!} x^2 + \frac{2^2(-n)(2-n)}{4!} x^4 + \frac{2^3(-n)(2-n)(4-n)}{6!} x^6 + \cdots , \\ y_2(x) & = x+\frac{2(1-n)}{3!} x^3 + \frac{2^2(1-n)(3-n)}{5!} x^5 + \frac{2^3(1-n)(3-n)(5-n)}{7!} x^7 + \cdots . \end{align*}

We then write

$y(x) = a_0 y_1(x) + a_1 y_2(x) .$

We also notice that if $$n$$ is a positive even integer, then $$y_1(x)$$ is a polynomial as all the coefficients in the series beyond a certain degree are zero. If $$n$$ is a positive odd integer, then $$y_2(x)$$ is a polynomial. For example, if $$n=4$$, then

$y_1(x) = 1 + \frac{2(-4)}{2!} x^2 + \frac{2^2(-4)(2-4)}{4!} x^4 = 1 - 4x^2 + \frac{4}{3} x^4 .$

### 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 $$k^{\text{th}}$$ coefficient.

Exercise 7.2.1: Use power series methods to solve $$y''+y = 0$$ at the point $$x_0 = 1$$.

Exercise 7.2.2: Use power series methods to solve $$y''+4xy = 0$$ at the point $$x_0 = 0$$.

Exercise 7.2.3: Use power series methods to solve $$y''-xy = 0$$ at the point $$x_0 = 1$$.

Exercise 7.2.4: Use power series methods to solve $$y''+x^2y = 0$$ at the point $$x_0 = 0$$.

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 $$x_0 = 0$$.

Exercise 7.2.6: Chebyshev's equation of order $$p$$ a) Solve $$(1-x^2)y''-xy' + p^2y = 0$$ using power series methods at $$x_0=0$$. b) For what $$p$$ is there a polynomial solution?

Exercise 7.2.7: Find a polynomial solution to $$(x^2+1) y''-2xy'+2y = 0$$ using power series methods.

Exercise 7.2.8: a) Use power series methods to solve $$(1-x)y''+y = 0$$ at the point $$x_0 = 0$$. b) Use the solution to part a) to find a solution for $$xy''+y=0$$ around the point $$x_0=1$$. \setcounter{exercise}{100}

Exercise 7.2.9: Use power series methods to solve $$y'' + 2 x^3 y = 0$$ at the point $$x_0 = 0$$.

\exsol{% %$%\begin{split} %0 = y''+2 x^3 y &= %\Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) %+ %2 x^3 %\Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) %\\ %&= %\Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) %+ %\Biggl( \sum_{k=0}^\infty 2 a_k x^{k+3} \Biggr) . %\\ %&= %\Biggl( \sum_{k=0}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) %+ %\Biggl( \sum_{k=3}^\infty 2 a_{k-3} x^k \Biggr) . %\\ %&= %2 a_2 + %6 a_3 x + %12 a_4 x^2 + %\Biggl( \sum_{k=3}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) %+ %\Biggl( \sum_{k=3}^\infty 2 a_{k-3} x^k \Biggr) . %\end{split} %$

$$a_2 = 0$$, $$a_3 = 0$$, $$a_4 = 0$$, recurrence relation (for $$k \geq 5): \(a_k = - 2 a_{k-5}$$, so:\\ $$y(x) = a_0 + a_1 x -2a_0 x^5 - 2 a_1 x^6 + 4 a_0 x^{10} + 4 a_1 x^{11} - 8 a_0 x^{15} - 8 a_1 x^{16} + \cdots$$ }

Exercise 7.2.10:[challenging] We can also use power series methods in nonhomogeneous equations. a) Use power series methods to solve $$y'' - x y = \frac{1}{1-x}$$ at the point $$x_0 = 0$$. Hint: Recall the geometric series. b) Now solve for the initial condition $$y(0)=0$$, $$y'(0) = 0$$. \exsol{% %$%\begin{split} %\frac{1}{1-x} = %\sum_{k=0}^\infty x^k %= %y''-x y &= %\Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) %- %x %\Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) %\\ %&= %\Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) %- %\Biggl( \sum_{k=0}^\infty a_k x^{k+1} \Biggr) . %\\ %&= %\Biggl( \sum_{k=0}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) %- %\Biggl( \sum_{k=1}^\infty a_{k-1} x^k \Biggr) . %\\ %&= %2 a_2 + %\Biggl( \sum_{k=1}^\infty (k+2)\,(k+1) \, a_{k+2} x^k \Biggr) %- %\Biggl( \sum_{k=1}^\infty a_{k-1} x^k \Biggr) . %\end{split} %$ a) $$a_2 = \frac{1}{2}$$, and for $$k \geq 1$$ we have $$a_k = a_{k-3} + 1$$, so \\ $$y(x) = a_0 + a_1 x + \frac{1}{2} x^2 + (a_0 + 1) x^3 + (a_1 + 1) x^4 + \frac{3}{2} x^5 + (a_0 + 2) x^6 + (a_1 + 2) x^7 + \frac{5}{2} x^8 + (a_0 + 3) x^9 + (a_1 + 3) x^{10} + \cdots$$ \\ b) $$y(x) = \frac{1}{2} x^2 + x^3 + x^4 + \frac{3}{2} x^5 + 2 x^6 + 2 x^7 + \frac{5}{2} x^8 + 3 x^9 + 3 x^{10} + \cdots$$ }

Exercise 7.2.1: Attempt to solve $$x^2 y'' - y = 0$$ at $$x_0 = 0$$ using the power series method of this section (\$x_0\) is a singular point). Can you find at least one solution? Can you find more than one solution?

$%\begin{split} %0 = x^2 y''-y &= %x^2 \Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^{k-2} \Biggr) %- %\Biggl( \sum_{k=0}^\infty a_k x^k \Biggr) %\\ %&= %\Biggl( \sum_{k=2}^\infty k\,(k-1) \, a_k x^k \Biggr) %- %a_0 %- %a_1 x %- %\Biggl( \sum_{k=2}^\infty a_k x^k \Biggr) . %\end{split} %$ %so $$a_0 = 0$$, $$a_1 = 0$$, $$k(k-1) a_k = a_k$$

Applying the method of this section directly we obtain $$a_k = 0$$ for all $$k$$ and so $$y(x) = 0$$ is the only solution we find.