3.3: Series Solutions Near an Ordinary Point II

In this section we continue to find series solutions

\begin{eqnarray*}
y=\sum_{n=0}^\infty a_n(x-x_0)^n
\end{eqnarray*}

of initial value problems

\begin{equation} \label{eq:3.3.1}
P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=a_0,\quad y'(x_0)=a_1,
\end{equation}

where \(P_0,P_1\), and \(P_2\) are polynomials and \(P_0(x_0)\ne0\), so \(x_0\) is an ordinary point of \eqref{eq:3.3.1}. However, here we consider cases where the differential equation in \eqref{eq:3.3.1} is not of the form

\begin{eqnarray*}
\left(1+\alpha(x-x_0)^2\right)y''+\beta(x-x_0) y'+\gamma y=0,
\end{eqnarray*}

so Theorem \((3.2.2)\) does not apply, and the computation of the coefficients \(\{a_n\}\) is more complicated. For the equations considered here it's difficult or impossible to obtain an explicit formula for \(a_n\) in terms of \(n\). Nevertheless, we can calculate as many coefficients as we wish. The next three examples illustrate this.

Example \(\PageIndex{1}\)

Find the coefficients \(a_0\), \(\dots\), \(a_7\) in the series solution \(y=\sum^\infty_{n=0} a_nx^n\) of the initial value problem

\begin{equation} \label{eq:3.3.2}
(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=-1,\quad y'(0)=-2.
\end{equation}

Answer

Here

\begin{eqnarray*}
Ly=(1+x+2x^2)y''+(1+7x)y'+2y.
\end{eqnarray*}

The zeros \((-1\pm i\sqrt7)/4\) of \(P_0(x)=1+x+2x^2\) have absolute value \(1/\sqrt2\), so Theorem \((3.2.2)\) implies that the series solution converges to the solution of \eqref{eq:3.3.2} on \((-1/\sqrt2,1/\sqrt2)\). Since

\begin{eqnarray*}
y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1}\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2},
\end{eqnarray*}

\begin{eqnarray*}
Ly&=&\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+\sum^\infty_{n=2}n(n-1)a_nx^{n-1} +2\sum^\infty_{n=2}n(n-1)a_nx^n\\
&&+\sum^\infty_{n=1}na_nx^{n-1}+7\sum^\infty_{n=1}na_nx^n+2\sum^\infty_{n=0} a_nx^n.
\end{eqnarray*}

Shifting indices so the general term in each series is a constant multiple of \(x^n\) yields

\begin{eqnarray*}
Ly&=&\sum^\infty_{n=0}(n+2)(n+1)a_{n+2}x^n+\sum^\infty_{n=0}(n+1)na_{n+1}x^n+2\sum^\infty_{n=0}n(n-1)a_nx^n\\
&&+\sum^\infty_{n=0}(n+1)a_{n+1}x^n+7\sum^\infty_{n=0}na_nx^n+ 2\sum^\infty_{n=0}a_nx^n =\sum^\infty_{n=0}b_nx^n,
\end{eqnarray*}

where

\begin{eqnarray*}
b_n=(n+2)(n+1)a_{n+2}+(n+1)^2a_{n+1}+(n+2)(2n+1)a_n.
\end{eqnarray*}

Therefore \(y=\sum^\infty_{n=0}a_nx^n\) is a solution of \(Ly=0\) if and only if

\begin{equation} \label{eq:3.3.3}
a_{n+2}=-{n+1\over n+2}\,a_{n+1}-{2n+1\over n+1}\,a_n,\,n\ge0.
\end{equation}

From the initial conditions in \eqref{eq:3.3.2}, \(a_0=y(0)=-1\) and \(a_1=y'(0)=-2\). Setting \(n=0\) in \eqref{eq:3.3.3} yields

\begin{eqnarray*}
a_2=-{1\over2}a_1-a_0=-{1\over2}(-2)-(-1)=2.
\end{eqnarray*}

Setting \(n=1\) in \eqref{eq:3.3.3} yields

\begin{eqnarray*}
a_3=-{2\over3}a_2-{3\over2}a_1=-{2\over3}(2)-{3\over2}(-2)={5\over3}.
\end{eqnarray*}

We leave it to you to compute \(a_4,a_5,a_6,a_7\) from \eqref{eq:3.3.3} and show that

\begin{eqnarray*}
y=-1-2x+2x^2+{5\over3}x^3-{55\over12}x^4+{3\over4}x^5+{61\over8}x^6- {443\over56}x^7+\cdots .
\end{eqnarray*}

We also leave it to you in Exercise \((3.3E.13)\) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of \eqref{eq:3.3.2} on \((-1/\sqrt2,1/\sqrt2)\).

Example \(\PageIndex{2}\)

Find the coefficients \(a_0\), \(\dots\), \(a_5\) in the series solution

\begin{eqnarray*}
y=\sum^\infty_{n=0} a_n(x+1)^n
\end{eqnarray*}

of the initial value problem

\begin{equation} \label{eq:3.3.4}
(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=2,\quad y'(-1)=-3.
\end{equation}

Answer

Since the desired series is in powers of \(x+1\) we rewrite the differential equation in \eqref{eq:3.3.4} as \(Ly=0\), with

\begin{eqnarray*}
Ly=\left(2+(x+1)\right)y''-\left(1-2(x+1)\right)y'-\left(3-(x+1)\right)y.
\end{eqnarray*}

Since

\begin{eqnarray*}
y=\sum^\infty_{n=0} a_n(x+1)^n,\quad y'=\sum^\infty_{n=1} n a_n(x+1)^{n-1}\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_n(x+1)^{n-2},
\end{eqnarray*}

\begin{eqnarray*}
Ly&=&2\sum^\infty_{n=2}n(n-1)a_n(x+1)^{n-2}+\sum^\infty_{n=2}n(n-1)a_n(x+1)^{n-1}\\
&&-\sum^\infty_{n=1}na_n(x+1)^{n-1}+2\sum^\infty_{n=1}na_n(x+1)^n\\
&&-3\sum^\infty_{n=0}a_n(x+1)^n+\sum_{n=0}^\infty a_n(x+1)^{n+1}.
\end{eqnarray*}

Shifting indices so that the general term in each series is a constant multiple of \((x+1)^n\) yields

\begin{eqnarray*}
Ly&=&2\sum^\infty_{n=0}(n+2)(n+1)a_{n+2}(x+1)^n+\sum^\infty_{n=0}(n+1)na_{n+1}
(x+1)^n\\[10pt]&&-\sum^\infty_{n=0}(n+1)a_{n+1}(x+1)^n
+\sum^\infty_{n=0}(2n-3)a_n(x+1)^n+\sum^\infty_{n=1}a_{n-1}(x+1)^n\\
&=&\sum^\infty_{n=0}b_n(x+1)^n,
\end{eqnarray*}

where

\begin{eqnarray*}
b_0=4a_2-a_1-3a_0
\end{eqnarray*}

and

\begin{eqnarray*}
b_n=2(n+2)(n+1)a_{n+2}+(n^2-1)a_{n+1}+(2n-3)a_n+a_{n-1},\quad n\ge1.
\end{eqnarray*}

Therefore \(y=\sum^\infty_{n=0}a_n(x+1)^n\) is a solution of \(Ly=0\) if and only if

\begin{equation} \label{eq:3.3.5}
a_2={1\over4}(a_1+3a_0)
\end{equation}

and

\begin{equation} \label{eq:3.3.6}
a_{n+2}=-{1\over2(n+2)(n+1)}\left[(n^2-1)a_{n+1}+(2n-3)a_n+a_{n-1}\right], \quad n\ge1.
\end{equation}

From the initial conditions in \eqref{eq:3.3.4}, \(a_0=y(-1)=2\) and \(a_1=y'(-1)=-3\). We leave it to you to compute \(a_2\), \(\dots\), \(a_5\) with \eqref{eq:3.3.5} and \eqref{eq:3.3.6} and show that the solution of \eqref{eq:3.3.4} is

\begin{eqnarray*}
y=-2-3(x+1)+{3\over4}(x+1)^2-{5\over12}(x+1)^3+{7\over48}(x+1)^4 -{1\over60}(x+1)^5+\cdots.
\end{eqnarray*}

We also leave it to you in Exercise \((3.3E.14)\) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of \eqref{eq:3.3.4} on the interval of convergence of the power series solution.

Example \(\PageIndex{3}\)

Find the coefficients \(a_0\), \(\dots\), \(a_5\) in the series solution \(y=\sum^\infty_{n=0} a_nx^n\) of the initial value problem

\begin{equation} \label{eq:3.3.7}
y''+3xy'+(4+2x^2)y=0,\quad y(0)=2,\quad y'(0)=-3.
\end{equation}

Answer

Here

\begin{eqnarray*}
Ly=y''+3xy'+(4+2x^2)y.
\end{eqnarray*}

Since

\begin{eqnarray*}
y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1},\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2},
\end{eqnarray*}

\begin{eqnarray*}
Ly&=&\sum^\infty_{n=2}n(n-1)a_nx^{n-2} +3\sum^\infty_{n=1}na_nx^n+4\sum^\infty_{n=0}a_nx^n+2\sum^\infty_{n=0} a_nx^{n+2}.
\end{eqnarray*}

Shifting indices so that the general term in each series is a constant multiple of \(x^n\) yields

\begin{eqnarray*}
Ly=\sum^\infty_{n=0}(n+2)(n+1)a_{n+2}x^n+\sum^\infty_{n=0}(3n+4)a_nx^n +2\sum^\infty_{n=2}a_{n-2}x^n=\sum_{n=0}^\infty b_nx^n
\end{eqnarray*}

where

\begin{eqnarray*}
b_0=2a_2+4a_0,\quad b_1=6a_3+7a_1,
\end{eqnarray*}

and

\begin{eqnarray*}
b_n=(n+2)(n+1)a_{n+2}+(3n+4)a_n+2a_{n-2},\quad n\ge2.
\end{eqnarray*}

Therefore \(y=\sum^\infty_{n=0}a_nx^n\) is a solution of \(Ly=0\) if and only if

\begin{equation} \label{eq:3.3.8}
a_2=-2a_0,\quad a_3=-{7\over6}a_1,
\end{equation}

and

\begin{equation} \label{eq:3.3.9}
a_{n+2}=-{1\over (n+2)(n+1)}\left[(3n+4)a_n+2a_{n-2}\right],\quad n\ge2.
\end{equation}

From the initial conditions in \eqref{eq:3.3.7}, \(a_0=y(0)=2\) and \(a_1=y'(0)=-3\). We leave it to you to compute \(a_2\), \(\dots\), \(a_5\) with \eqref{eq:3.3.8} and \eqref{eq:3.3.9} and show that the solution of \eqref{eq:3.3.7} is

\begin{eqnarray*}
y=2-3x-4x^2+{7\over2}x^3+3x^4-{79\over40}x^5+\cdots.
\end{eqnarray*}

We also leave it to you in Exercise \((3.3E.15)\) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of \eqref{eq:3.3.9} on the interval of convergence of the power series solution.