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\begin{document}


\section{Power Series Solutions\label{Power Series Solutions}}

Consider the second-order homogeneous linear differential equation 
\[
a_{0}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{2}(x)y=0\text{.} 
\]

The methods presented in this section are used to find nontrivial solutions
that are expressible in the form of a power series. In many cases, the
methods yield two linearly independent solutions. If, however, only one such
solution is found, then another linearly independent solution can be found
by a \hyperref{reduction of order}{}{}{ODE02_01.TEX#Reduction of Order}.

\begin{definition}
A function $f$ is \textbf{analytic\label{Analytic}} at a real number $x_{0}$
if there exists a power series about $x_{0}$ ,\textbf{\ } 
\[
\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n},
\]
which converges to $f(x)$ for all $\ x$ in some open interval containing $%
x_{0}$.
\end{definition}

\begin{theorem}
To each power series $\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}$ there
corresponds a number $R$ , where $0\leq R\leq \infty $, such that the series
converges in the interval $\left| x-x_{0}\right| <R$ and diverges whenever $%
\left| x-x_{0}\right| >R$. The number $R$ is called the \textbf{radius of
convergence\label{Radius of Convergence}} and is given by 
\[
\frac{1}{R}=\lim \sup_{n->\infty }\left| c_{n}\right| ^{\frac{1}{n}}\text{.}
\]
\end{theorem}

\begin{theorem}
Suppose $f\left( x\right) =\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}$ converges
for all $x$ in some open interval $I$ containing $x_{0}$.
\end{theorem}

\begin{enumerate}
\begin{enumerate}
\begin{enumerate}
\item  The function $f$ is infinitely differentiable on $I$ and the
derivatives of $f$ have the power series expansions 
\[
f^{(k)}(x)=\sum_{n=k}^{\infty }n(n-1)\cdots (n-k+1)c_{n}(x-x_{0})^{n-k}\text{
}(k=1,2,\ldots ;x\in I)\text{ }.
\]

\item  The coefficients $c_{n}$ of the series are given by 
\[
c_{n}=\frac{f^{(n)}(x_{0})}{n!}\text{ for all }n\text{.}
\]

\item  If $f(x)=0$ for all $x\in I$, then $c_{n}=0$ for all $n.$
\end{enumerate}
\end{enumerate}
\end{enumerate}

Dividing both sides of the differential equation by \ $a_{0}(x)$ leads to
the normalized equation 
\[
\frac{d^{2}y}{dx^{2}}+P_{1}(x)\frac{dy}{dx}+P_{2}(x)y=0 
\]

where 
\[
P_{1}(x)=\frac{a_{1}(x)}{a_{0}(x)}\text{ and }P_{2}(x)=\frac{a_{2}(x)}{%
a_{0}(x)}\text{.} 
\]

\begin{definition}
A point $x_{0}$ is called an \textbf{ordinary point\label{Ordinary Point}}
of the differential equation if the functions $P_{1}$ and $P_{2}$ are both
analytic at $x_{0}$. If either of these functions is not analytic at $x_{0}$%
, then $x_{0}$ is called a \textbf{singular point\label{Singular Point}} of
the differential equation.
\end{definition}

\subsection{\protect\vspace{1pt}Solutions About an Ordinary Point}

\begin{theorem}
If $x_{0}$ is an ordinary point, then the equation has two linearly
independent power series solutions of the form 
\[
\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}\text{.}
\]
If the power series expansions for the functions $P_{1}$ and $P_{2}$
converge in an interval $I$ containing $x_{0}$, then this series converges
in $I$.
\end{theorem}

\begin{enumerate}
\item  Assume a solution has the form 
\[
y=\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}\text{.}
\]
Differentiate this series term by term to obtain expansions for the first
and second derivatives of $y$%
\begin{eqnarray*}
\frac{dy}{dx} &=&\sum_{n=1}^{\infty }nc_{n}(x-x_{0})^{n-1} \\
\frac{d^{2}y}{dx^{2}} &=&\sum_{n=2}^{\infty }n(n-1)c_{n}(x-x_{0})^{n-2}\text{%
.}
\end{eqnarray*}

\item  Substitute the three series in step 1. into the differential equation
and simplify so that it takes the form 
\[
d_{0}+d_{1}(x-x_{0})+d_{2}(x-x_{0})^{2}+\cdots =0\text{.}
\]
Each $d_{k}$ is a function of certain coefficients $c_{n}$ of the series.

\item  Since the left side of the equation in step 2 is assumed to converge
in some interval containing $x_{0}$, it follows that 
\[
d_{0}=d_{1}=d_{2}=\cdots =0\text{.}
\]
These equations give conditions that the coefficients $c_{n}$ must satisfy
for the series to be a solution.
\end{enumerate}

Two arbitrary constants appear when solving for the coefficients $c_{n}$.
When the coefficients are substituted into the series, the series will
separate into a linear combination of two linearly independent solutions
with the arbitrary constants as the coefficients of the combination.

\vspace{1pt}

\subsection{Solutions About a Singular Point}

\begin{definition}
Suppose $x_{0}$ is a singular point. Let $P_{1}$ and $P_{2}$ be the
functions in the normalized equation. If the functions 
\[
(x-x_{0})P_{1}(x)\text{ and }(x-x_{0})^{2}P_{2}(x)
\]
are both analytic at $x_{0}$, then $x_{0}$ is called a \textbf{regular
singular point\label{Regular Singular Point} }of the differential equation.
If either function is not analytic at $x_{0}$, then $x_{0}$ is called an 
\textbf{irregular singular point\label{Irregular Singular Point} }of the
differential equation.
\end{definition}

\begin{theorem}
Let $x_{0}$ be a regular singular point. Then there is at least one
nontrivial solution of the form 
\[
(x-x_{0})^{r}\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}
\]
where $r$ is a complex constant. If the functions $(x-x_{0})P_{1}(x)$ and $%
(x-x_{0})^{2}P_{2}(x)$ converge in an interval $I$ containing $x_{0}$, then
the series converges in $I$, except possibly at $x_{0}$.
\end{theorem}

\subsubsection{The Method of Frobenius\label{Method of Frobenius}}

\begin{enumerate}
\item  Assume a solution has the form 
\[
y=\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n+r}\text{.}
\]
where $c_{0}\neq 0$. Differentiate this series term by term to obtain series
expansions for the first and second derivatives of $y$%
\begin{eqnarray*}
\frac{dy}{dx} &=&\sum_{n=0}^{\infty }(n+r)c_{n}(x-x_{0})^{n+r-1} \\
\frac{d^{2}y}{dx^{2}} &=&\sum_{n=0}^{\infty
}(n+r)(n+r-1)c_{n}(x-x_{0})^{n+r-2}\text{.}
\end{eqnarray*}

\item  Substitute the three series in step 1. into the differential equation
and simplify so that it takes the form 
\[
d_{0}+d_{1}(x-x_{0})+d_{2}(x-x_{0})^{2}+\cdots =0\text{.}
\]
Each $d_{k}$ is a function of $r$ and certain of the coefficients $c_{n}$ of
the series for $y$.

\item  Since the left side of the equation in step 2 is assumed to converge
in some interval containing $x_{0}$, it follows that 
\[
d_{0}=d_{1}=d_{2}=\cdots =0\text{.}
\]
It can be shown that $d_{0}=0$ yields a quadratic equation in $r$, called
the \textbf{indicial equation\label{Indicial Equation}} associated with the
point $x_{0}$. The two roots $r_{1}$ and $r_{2}$ (where $\func{Re}r_{1}\geq 
\func{Re}r_{2}$) of the indicial equation are called the \textbf{exponents}
of the differential equation. The remaining equations lead to a set of
conditions that the coefficients $c_{n}$ must satisfy in order that the
series for $y$ is a solution.

\item  Substitute the root $r_{1}$ of the indicial equation into the set of
conditions obtained in step 3. and choose the $c_{n}$ to satisfy these
conditions. The resulting series with $r=r_{1}$ is a solution.

\item  Suppose $r_{1}-r_{2}$ is neither zero nor a positive integer. Proceed
as in step 4. with $r_{2}$ instead of $r_{1}$ and obtain another solution
which is linearly independent of the solution obtained using $r_{1}$. If $%
r_{1}-r_{2}$ is zero or a positive integer, another linearly independent
solution is obtained by a reduction of order.
\end{enumerate}

\begin{theorem}
Suppose $x_{0}$ is a regular singular point. Let $r_{1}$ and $r_{2}$ (where $%
\func{Re}r_{1}\geq \func{Re}r_{2})$ be the roots of the indicial equation
associated with the point $x_{0}$.
\end{theorem}

\begin{enumerate}
\begin{enumerate}
\begin{enumerate}
\item  If $r_{1}-r_{2}$ is neither zero nor a positive integer, then there
are two linearly independent solutions 
\[
y_{1}(x)=(x-x_{0})^{r_{1}}\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}
\]
and 
\[
y_{2}(x)=(x-x_{0})^{r_{2}}\sum_{n=0}^{\infty }d_{n}(x-x_{0})^{n}
\]
where $c_{0}\neq 0,d_{0}\neq 0$.

\item  If $r_{1}-r_{2}$ is either zero or a positive integer, then there are
two linearly independent solutions 
\[
y_{1}(x)=(x-x_{0})^{r_{1}}\sum_{n=0}^{\infty }c_{n}(x-x_{0})^{n}\text{ }
\]
where $c_{0}\neq 0$, and 
\[
y_{2}(x)=(x-x_{0})^{r_{2}}\sum_{n=0}^{\infty
}d_{n}(x-x_{0})^{n}+Cy_{1}(x)\ln \left| x-x_{0}\right| 
\]
where $d_{0}\neq 0$ and $C$ is a constant.
\end{enumerate}
\end{enumerate}
\end{enumerate}

\begin{theorem}
Suppose $x_{0}$ is a regular singular point. Let $P_{1}$ and $P_{2}$ be the
functions in the normalized equation. Then the indicial equation associated
with $x_{0}$ is given by 
\[
r^{2}+(a-1)r+b=0
\]
where 
\[
a=\lim_{x\rightarrow x_{0}}(x-x_{0})P_{1}(x)
\]
and 
\[
b=\lim_{x\rightarrow x_{0}}(x-x_{0})^{2}P_{2}(x)\text{.}
\]
\end{theorem}

\subsection{\ Related Scientific Notebook Topics}

{\small \hyperref{Solve ODE, series}{}{}{../../Help/DM10-1.tex#Solve ODE,
series}}

\noindent \hrulefill

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