# 7.3: Series Solutions Near an Ordinary Point II

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In this section we continue to find series solutions

$y=\sum_{n=0}^\infty a_n(x-x_0)^n \nonumber$

of initial value problems

$\label{eq:7.3.1} P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=a_0,\quad y'(x_0)=a_1,$

where $$P_0,P_1$$, and $$P_2$$ are polynomials and $$P_0(x_0)\ne0$$, so $$x_0$$ is an ordinary point of Equation \ref{eq:7.3.1}. However, here we consider cases where the differential equation in Equation \ref{eq:7.3.1} is not of the form

$\left(1+\alpha(x-x_0)^2\right)y''+\beta(x-x_0) y'+\gamma y=0,\nonumber$

so Theorem 7.2.2 does not apply, and the computation of the coefficients $$\{a_n\}$$ is more complicated. For the equations considered here it is 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 7.4.1

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

$\label{eq:7.3.2} (1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=-1,\quad y'(0)=-2.$

###### Solution

Here

$Ly=(1+x+2x^2)y''+(1+7x)y'+2y.\nonumber$

The zeros $$(-1\pm i\sqrt7)/4$$ of $$P_0(x)=1+x+2x^2$$ have absolute value $$1/\sqrt2$$, so Theorem 7.2.2 implies that the series solution converges to the solution of Equation \ref{eq:7.3.2} on $$(-1/\sqrt2,1/\sqrt2)$$. Since

$y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1}\quad \text{and}\quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2},\nonumber$

\begin{aligned} 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\\[4pt] &+\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{aligned}

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

\begin{aligned} 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\\[10pt] &+\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{aligned}

where

$b_n=(n+2)(n+1)a_{n+2}+(n+1)^2a_{n+1}+(n+2)(2n+1)a_n.\nonumber$

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

$\label{eq:7.3.3} a_{n+2}=-{n+1\over n+2}\,a_{n+1}-{2n+1\over n+1}\,a_n,\,n\ge0.$

From the initial conditions in Equation \ref{eq:7.3.2}, $$a_0=y(0)=-1$$ and $$a_1=y'(0)=-2$$. Setting $$n=0$$ in Equation \ref{eq:7.3.3} yields

$a_2=-{1\over2}a_1-a_0=-{1\over2}(-2)-(-1)=2.\nonumber$

Setting $$n=1$$ in Equation \ref{eq:7.3.3} yields

$a_3=-{2\over3}a_2-{3\over2}a_1=-{2\over3}(2)-{3\over2}(-2)={5\over3}.\nonumber$

We leave it to you to compute $$a_4,a_5,a_6,a_7$$ from Equation \ref{eq:7.3.3} and show that

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

We also leave it to you (Exercise [exer:7.3.13}) to verify numerically that the Taylor polynomials $$T_N(x)=\sum_{n=0}^Na_nx^n$$ converge to the solution of Equation \ref{eq:7.3.2} on $$(-1/\sqrt2,1/\sqrt2)$$.

## Example 7.4.2

Find the coefficients $$a_0$$, …, $$a_5$$ in the series solution

$y=\sum^\infty_{n=0} a_n(x+1)^n\nonumber$

of the initial value problem

$\label{eq:7.3.4} (3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=2,\quad y'(-1)=-3.$

Since the desired series is in powers of $$x+1$$ we rewrite the differential equation in Equation \ref{eq:7.3.4} as $$Ly=0$$, with

$Ly=\left(2+(x+1)\right)y''-\left(1-2(x+1)\right)y'-\left(3-(x+1)\right)y.\nonumber$

Since

$y=\sum^\infty_{n=0} a_n(x+1)^n,\quad y'=\sum^\infty_{n=1} n a_n(x+1)^{n-1}\quad \text{and}\quad y''=\sum^\infty_{n=2}n(n-1)a_n(x+1)^{n-2},\nonumber$

\begin{aligned} 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\\[4pt] &-3\sum^\infty_{n=0}a_n(x+1)^n+\sum_{n=0}^\infty a_n(x+1)^{n+1}.\end{aligned}

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

\begin{aligned} 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\\[10pt] &= \sum^\infty_{n=0}b_n(x+1)^n,\end{aligned}

where

$b_0=4a_2-a_1-3a_0\nonumber$

and

$b_n=2(n+2)(n+1)a_{n+2}+(n^2-1)a_{n+1}+(2n-3)a_n+a_{n-1},\quad n\ge1.\nonumber$

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

$\label{eq:7.3.5} a_2={1\over4}(a_1+3a_0)$

and

$\label{eq:7.3.6} a_{n+2}=-{1\over2(n+2)(n+1)}\left[(n^2-1)a_{n+1}+(2n-3)a_n+a_{n-1}\right], \quad n\ge1.$

From the initial conditions in Equation \ref{eq:7.3.4}, $$a_0=y(-1)=2$$ and $$a_1=y'(-1)=-3$$. We leave it to you to compute $$a_2$$, …, $$a_5$$ with Equation \ref{eq:7.3.5} and Equation \ref{eq:7.3.6} and show that the solution of Equation \ref{eq:7.3.4} is

$y=-2-3(x+1)+{3\over4}(x+1)^2-{5\over12}(x+1)^3+{7\over48}(x+1)^4 -{1\over60}(x+1)^5+\cdots.\nonumber$

We also leave it to you (Exercise [exer:7.3.14}) to verify numerically that the Taylor polynomials $$T_N(x)=\sum_{n=0}^Na_nx^n$$ converge to the solution of Equation \ref{eq:7.3.4} on the interval of convergence of the power series solution.

## Example 7.4.3

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

$\label{eq:7.3.7} y''+3xy'+(4+2x^2)y=0,\quad y(0)=2,\quad y'(0)=-3.$

###### Solution

Here

$Ly=y''+3xy'+(4+2x^2)y.\nonumber$

Since

$y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1},\quad\text {and} \quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2},\nonumber$

\begin{aligned} 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{aligned}

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

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

where

$b_0=2a_2+4a_0,\quad b_1=6a_3+7a_1,\nonumber$

and

$b_n=(n+2)(n+1)a_{n+2}+(3n+4)a_n+2a_{n-2},\quad n\ge2.\nonumber$

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

$\label{eq:7.3.8} a_2=-2a_0,\quad a_3=-{7\over6}a_1,$

and

$\label{eq:7.3.9} a_{n+2}=-{1\over (n+2)(n+1)}\left[(3n+4)a_n+2a_{n-2}\right],\quad n\ge2.$

From the initial conditions in Equation \ref{eq:7.3.7}, $$a_0=y(0)=2$$ and $$a_1=y'(0)=-3$$. We leave it to you to compute $$a_2$$, …, $$a_5$$ with Equation \ref{eq:7.3.8} and Equation \ref{eq:7.3.9} and show that the solution of Equation \ref{eq:7.3.7} is

$y=2-3x-4x^2+{7\over2}x^3+3x^4-{79\over40}x^5+\cdots.\nonumber$

We also leave it to you (Exercise [exer:7.3.15}) to verify numerically that the Taylor polynomials $$T_N(x)=\sum_{n=0}^Na_nx^n$$ converge to the solution of Equation \ref{eq:7.3.9} on the interval of convergence of the power series solution.

This page titled 7.3: Series Solutions Near an Ordinary Point II is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.