are defined for \(x\) near \(x_0\text{.}\) If \(p(x_0) = 0\text{,}\) 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.

Example7.2.1

Let us start with a very simple example

\begin{equation*}
y'' - y = 0 .
\end{equation*}

Let us try a power series solution near \(x_0 = 0\text{,}\) 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

\begin{equation*}
y = \sum_{k=0}^\infty a_k x^k .
\end{equation*}

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

\begin{equation*}
y' = \sum_{k=1}^\infty k a_k x^{k-1} .
\end{equation*}

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

The above equation is called a 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 \(a_1\text{,}\) 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

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

\begin{equation*}
y =
a_0 \cosh x + a_1 \sinh x .
\end{equation*}

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.

Example7.2.2

Let us do a more complex example. Suppose we wish to solve Airy's equation^{ 1 }Named after the English mathematician Sir George Biddell Airy (1801–1892)., that is

\begin{equation*}
y'' - xy = 0 ,
\end{equation*}

near the point \(x_0 = 0\text{.}\) Note that \(x_0 = 0\) is an ordinary point.

We try

\begin{equation*}
y = \sum_{k=0}^\infty a_k x^k .
\end{equation*}

Now we jump in steps of three. First we notice that since \(a_2 = 0\) we must have that, \(a_5 = 0\text{,}\) \(a_8 = 0\text{,}\) \(a_{11}=0\text{,}\) etc. In general \(a_{3n+2} = 0\text{.}\)

The constants \(a_0\) and \(a_1\) are arbitrary and we obtain

and write the general solution to the equation as \(y(x)= a_0 y_1(x) + a_1 y_2(x)\text{.}\) Notice from the power series that \(y_1(0) = 1\) and \(y_2(0) = 0\text{.}\) Also, \(y_1'(0) = 0\) and \(y_2'(0) = 1\text{.}\) Therefore \(y(x)\) is a solution that satisfies the initial conditions \(y(0) = a_0\) and \(y'(0) = a_1\text{.}\)

The functions \(y_1\) and \(y_2\) cannot be written in terms of the elementary functions that you know. See Figure 8.2.3 for the plot of the solutions \(y_1\) and \(y_2\text{.}\) 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 \(y''-y=0\)).

Sometimes a solution may turn out to be a polynomial.

Example7.2.3

Let us find a solution to the so-called Hermite's equation of order \(n\)^{ 2 }Named after the French mathematician Charles Hermite (1822–1901). is the equation

\begin{equation*}
y'' -2xy' + 2n y = 0 .
\end{equation*}

Let us find a solution around the point \(x_0 = 0\text{.}\) We try

\begin{equation*}
y = \sum_{k=0}^\infty a_k x^k .
\end{equation*}

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 \(n=4\text{,}\) then

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.

Exercise7.2.1

Use power series methods to solve \(y''+y = 0\) at the point \(x_0 = 1\text{.}\)

Exercise7.2.2

Use power series methods to solve \(y''+4xy = 0\) at the point \(x_0 = 0\text{.}\)

Exercise7.2.3

Use power series methods to solve \(y''-xy = 0\) at the point \(x_0 = 1\text{.}\)

Exercise7.2.4

Use power series methods to solve \(y''+x^2y = 0\) at the point \(x_0 = 0\text{.}\)

Exercise7.2.5

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 \(x_0 = 0\text{.}\)

Exercise7.2.6Chebyshev's equation of order \(p\)

a) Solve \((1-x^2)y''-xy' + p^2y = 0\) using power series methods at \(x_0=0\text{.}\) b) For what \(p\) is there a polynomial solution?

Exercise7.2.7

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

Exercise7.2.8

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

Exercise7.2.101

Use power series methods to solve \(y'' + 2 x^3 y = 0\) at the point \(x_0 =
0\text{.}\)

(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 \(x_0 = 0\text{.}\) Hint: Recall the geometric series. b) Now solve for the initial condition \(y(0)=0\text{,}\) \(y'(0) = 0\text{.}\)

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?