7.3 Series Solutions Near an Ordinary Point II

In this section we continue to find series solutions

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

of initial value problems

\begin{equation} 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, \tag{7.3.1}\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 (7.3.1). However, here we consider cases where the differential equation in (7.3.1) is not of the form

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

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’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 7.3.1

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

\begin{equation} (1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=-1,\quad y'(0)=-2. \tag{7.3.2}\end{equation}

Solution Here

\[ Ly=(1+x+2x^2)y''+(1+7x)y'+2y. \]

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 (7.3.2) on \((-1/\sqrt2,1/\sqrt2)\). Since

\[ y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1}\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2}, \]

\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 \\[10pt] &&+\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 \\[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{eqnarray*}

where

\[ b_n=(n+2)(n+1)a_{n+2}+(n+1)^2a_{n+1}+(n+2)(2n+1)a_n. \]

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

\begin{equation} a_{n+2}=-{n+1\over n+2}\,a_{n+1}-{2n+1\over n+1}\,a_n,\,n\ge0. \tag{7.3.3}\end{equation}

From the initial conditions in (7.3.2), \(a_0=y(0)=-1\) and \(a_1=y'(0)=-2\). Setting \(n=0\) in (7.3.3) yields

\[ a_2=-{1\over2}a_1-a_0=-{1\over2}(-2)-(-1)=2. \]

Setting \(n=1\) in (7.3.3) yields

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

We leave it to you to compute \(a_4,a_5,a_6,a_7\) from (7.3.3) and show that

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

We also leave it to you (Exercise 13) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of (7.3.2) on \((-1/\sqrt2,1/\sqrt2)\).

Example 7.3.2

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

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

of the initial value problem

\begin{equation} (3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=2,\quad y'(-1)=-3. \tag{7.3.4}\end{equation}

Solution Since the desired series is in powers of \(x+1\) we rewrite the differential equation in (7.3.4) as \(Ly=0\), with

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

Since

\[ y=\sum^\infty_{n=0} a_n(x+1)^n,\quad y'=\sum^\infty_{n=1} n a_n(x+1)^{n-1}\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_n(x+1)^{n-2}, \]

\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 \\[6pt] &&-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 \\[10pt] &=&\sum^\infty_{n=0}b_n(x+1)^n, \end{eqnarray*}

where

\[ b_0=4a_2-a_1-3a_0 \]

and

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

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

\begin{equation} a_2={1\over4}(a_1+3a_0) \tag{7.3.5}\end{equation}

and

\begin{equation} 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. \tag{7.3.6}\end{equation}

From the initial conditions in (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 (7.3.5) and (7.3.6) and show that the solution of (7.3.4) is

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

We also leave it to you (Exercise 14) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of (7.3.4) on the interval of convergence of the power series solution.

Example 7.3.3

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

\begin{equation} y''+3xy'+(4+2x^2)y=0,\quad y(0)=2,\quad y'(0)=-3. \tag{7.3.7}\end{equation}

Solution Here

\[ Ly=y''+3xy'+(4+2x^2)y. \]

Since

\[ y=\sum^\infty_{n=0} a_nx^n,\quad y'=\sum^\infty_{n=1} n a_nx^{n-1},\quad\mbox{ and }\quad y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2}, \]

\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

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

where

\[ b_0=2a_2+4a_0,\quad b_1=6a_3+7a_1, \]

and

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

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

\begin{equation} a_2=-2a_0,\quad a_3=-{7\over6}a_1, \tag{7.3.8}\end{equation}

and

\begin{equation} a_{n+2}=-{1\over (n+2)(n+1)}\left[(3n+4)a_n+2a_{n-2}\right],\quad n\ge2. \tag{7.3.9}\end{equation}

From the initial conditions in (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 (7.3.8) and (7.3.9) and show that the solution of (7.3.7) is

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

We also leave it to you (Exercise 15) to verify numerically that the Taylor polynomials \(T_N(x)=\sum_{n=0}^Na_nx^n\) converge to the solution of (7.3.9) on the interval of convergence of the power series solution.

7.3 Exercises

In Exercises 1–12 find the coefficients \(a_0\),…, \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

C \((1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3\)

Show answer

\(y=\dst{2-3x-2x^2+{7\over2}x^3-{55\over12}x^4+{59\over8}x^5-{83\over6}x^6 +{9547\over336}x^7+\cdots}\)
C \((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)

Show answer

\(y=\dst{-1+2x-4x^3+4x^4+4x^5-12x^6+4x^7+\cdots}\)
C \((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)

Show answer

\(y=\dst{1+x^2-{2\over3}x^3+{11\over6}x^4-{9\over5}x^5+ {329\over90}x^6-{1301\over315}x^7+\cdots}\)
C \((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)

Show answer

\(y=\dst{x-x^2-{7\over2}x^3+{15\over2}x^4+{45\over8}x^5 -{261\over8}x^6+{207\over16}x^7+\cdots}\)
C \((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)

Show answer

\(y=\dst{4+3x-{15\over4}x^2+{1\over4}x^3+{11\over16}x^4-{5\over16}x^5 +{1\over20}x^6+{1\over120}x^7+\cdots}\)
C \((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)

Show answer

\(y=\dst{7+3x-{16\over3}x^2+{13\over3}x^3-{23\over9}x^4+{10\over9}x^5 -{7\over27}x^6-{1\over9}x^7+\cdots}\)
C \((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)

Show answer

\(y=\dst{2+5x-{7\over4}x^2-{3\over16}x^3+{37\over192}x^4 -{7\over192}x^5-{1\over1920}x^6+{19\over11520}x^7+\cdots}\)
C \((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)

Show answer

\(y=\dst{1-(x-1)+{4\over3}(x-1)^3-{4\over3}(x-1)^4-{4\over5}(x-1)^5 +{136\over45}(x-1)^6-{104\over63}(x-1)^7+\cdots}\)
C \((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)

Show answer

\(y=\dst{1-(x+1)+4(x+1)^2-{13\over3}(x+1)^3+{77\over6}(x+1)^4 -{278\over15}(x+1)^5+{1942\over45}(x+1)^6-{23332\over315}(x+1)^7+\cdots}\)
C \((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)

Show answer

\(y=\dst{2-(x-1)-{1\over2}(x-1)^2+{5\over3}(x-1)^3-{19\over12}(x-1)^4 +{7\over30}(x-1)^5+{59\over45}(x-1)^6-{1091\over630}(x-1)^7+\cdots}\)
C \((2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3\)

Show answer

\(y=\dst{-2+3(x+1)-{1\over2}(x+1)^2-{2\over3}(x+1)^3+{5\over8}(x+1)^4 -{11\over30}(x+1)^5+{29\over144}(x+1)^6-{101\over840}(x+1)^7+\cdots}\)
C \(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)

Show answer

\(y=\dst{1-2(x-1)-3(x-1)^2+8(x-1)^3-4(x-1)^4-{42\over5}(x-1)^5 +19(x-1)^6-{604\over35}(x-1)^7+\cdots}\)
L Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).
1. Use differential equations software to solve the initial value problem
  
  \[ (1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag*{\rm(A)} \]
  
  numerically on \((-r,r)\). (See Example 7.3.1.)
2. For \(N=2\), \(3\), \(4\), …, compute \(a_2\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A), and graph
  
  \[ T_N(x)=\sum_{n=0}^N a_nx^n \]
  
  and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.
L Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<2\).
1. Use differential equations software to solve the initial value problem
  
  \[ (3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \tag*{\rm(A)} \]
  
  numerically on \((-1-r,-1+r)\). (See Example 7.3.2. Why this interval?)
2. For \(N=2\), \(3\), \(4\), …, compute \(a_2,\dots,a_N\) in the power series solution
  
  \[ y=\sum_{n=0}^\infty a_n(x+1)^n \]
  
  of (A), and graph
  
  \[ T_N(x)=\sum_{n=0}^N a_n(x+1)^n \]
  
  and the solution obtained in (a) on \((-1-r,-1+r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.
L Do the following experiment for several choices of \(a_0\), \(a_1\), and \(r\), with \(r>0\).
1. Use differential equations software to solve the initial value problem
  
  \[ y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag*{\rm(A)} \]
  
  numerically on \((-r,r)\). (See Example 7.3.3.)
2. Find the coefficients \(a_0\), \(a_1\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of (A), and graph
  
  \[ T_N(x)=\sum_{n=0}^N a_nx^n \]
  
  and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs.
L Do the following experiment for several choices of \(a_0\) and \(a_1\).
1. Use differential equations software to solve the initial value problem
  
  \[ (1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag*{\rm(A)} \]
  
  numerically on \((-r,r)\).
2. Find the coefficients \(a_0\), \(a_1\), …, \(a_N\) in the power series solution \(y=\sum_{n=0}^Na_nx^n\) of (A), and graph
  
  \[ T_N(x)=\sum_{n=0}^N a_nx^n \]
  
  and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there’s no perceptible difference between the two graphs. What happens as you let \(r\to1\)?
L Follow the directions of Exercise 16 for the initial value problem

\[ (1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1. \]
L Follow the directions of Exercise 16 for the initial value problem

\[ (1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1. \]

In Exercises 19–28 find the coefficients \(a_0\), …, \(a_N\) for \(N\) at least \(7\) in the series solution

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

of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

C \((2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7\)

Show answer

\(y=\dst{2-7x-4x^2-{17\over6}x^3-{3\over4}x^4-{9\over40}x^5+\cdots}\)
C \((1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2\)

Show answer

\(y=\dst{1-2(x-1)+{1\over2}(x-1)^2-{1\over6}(x-1)^3+{5\over36}(x-1)^4 -{73\over1080}(x-1)^5+\cdots}\)
C \((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

Show answer

\(y=\dst{2-(x+2)-{7\over2}(x+2)^2+{4\over3}(x+2)^3-{1\over24}(x+2)^4 +{1\over60}(x+2)^5+\cdots}\)
C \((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)

Show answer

\(y=\dst{2-2(x+3)-(x+3)^2+(x+3)^3-{11\over12}(x+3)^4+ {67\over60}(x+3)^5+\cdots}\)
C \((2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2\)

Show answer

\(y=\dst{-1+2x+{1\over3}x^3-{5\over12}x^4+{2\over5}x^5+\cdots}\)
C \((3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3\)

Show answer

\(y=\dst{2-3(x+1)+{7\over2}(x+1)^2-5(x+1)^3+{197\over24}(x+1)^4 -{287\over20}(x+1)^5+\cdots}\)
C \((3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3\)

Show answer

\(y=\dst{-2+3(x+2)-{9\over2}(x+2)^2+{11\over6}(x+2)^3+{5\over24}(x+2)^4 +{7\over20}(x+2)^5+\cdots}\)
C \((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)

Show answer

\(y=\dst{2-4(x-2)-{1\over2}(x-2)^2+{2\over9}(x-2)^3+{49\over432}(x-2)^4 +{23\over1080}(x-2)^5+\cdots}\)
C \((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)

Show answer

\(y=\dst{1+2(x+4)-{1\over6}(x+4)^2-{10\over27}(x+4)^3+{19\over648}(x+4)^4 +{13\over324}(x+4)^5+\cdots}\)
C \((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)

Show answer

\(y=\dst{-1+2(x+1)-{1\over4}(x+1)^2+{1\over2}(x+1)^3-{65\over96}(x+1)^4 +{67\over80}(x+1)^5+\cdots}\)
Show that the coefficients in the power series in \(x\) for the general solution of

\[ (1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0 \]

satisfy the recurrrence relation

\[ a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n. \]
1. Let \(\alpha\) and \(\beta\) be constants, with \(\beta\ne0\). Show that \(y=\sum_{n=0}^\infty a_nx^n\) is a solution of
  
  \[ (1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \tag*{\rm (A)} \]
  
  if and only if
  
  \[ a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \tag*{\rm (B)} \]
  
  An equation of this form is called a second order homogeneous linear difference equation. The polynomial \(p(r)=r^2+\alpha r+\beta\) is called the characteristic polynomial of (B). If \(r_1\) and \(r_2\) are the zeros of \(p\), then \(1/r_1\) and \(1/r_2\) are the zeros of
  
  \[ P_0(x)=1+\alpha x+\beta x^2. \]
2. Suppose \(p(r)=(r-r_1)(r-r_2)\) where \(r_1\) and \(r_2\) are real and distinct, and let \(\rho\) be the smaller of the two numbers \(\{1/|r_1|,1/|r_2|\}\). Show that if \(c_1\) and \(c_2\) are constants then the sequence
  
  \[ a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0 \]
  
  satisfies (B). Conclude from this that any function of the form
  
  \[ y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n \]
  
  is a solution of (A) on \((-\rho,\rho)\).
3. Use (b) and the formula for the sum of a geometric series to show that the functions
  
  \[ y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x} \]
  
  form a fundamental set of solutions of (A) on \((-\rho,\rho)\).
4. Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on any interval that does’nt contain either \(1/r_1\) or \(1/r_2\).
5. Suppose \(p(r)=(r-r_1)^2\), and let \(\rho=1/|r_1|\). Show that if \(c_1\) and \(c_2\) are constants then the sequence
  
  \[ a_n=(c_1+c_2n)r_1^n,\quad n\ge0 \]
  
  satisfies (B). Conclude from this that any function of the form
  
  \[ y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n \]
  
  is a solution of (A) on \((-\rho,\rho)\).
6. Use (e) and the formula for the sum of a geometric series to show that the functions
  
  \[ y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2} \]
  
  form a fundamental set of solutions of (A) on \((-\rho,\rho)\).
7. Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on any interval that does not contain \(1/r_1\).
Use the results of Exercise 30 to find the general solution of the given equation on any interval on which polynomial multiplying \(y''\) has no zeros.

(a) \((1+3x+2x^2)y''+(6+8x)y'+4y=0\)

(b) \((1-5x+6x^2)y''-(10-24x)y'+12y=0\)

(c) \((1-4x+4x^2)y''-(8-16x)y'+8y=0\)

(d) \((4+4x+x^2)y''+(8+4x)y'+2y=0\)

(e) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

Show answer

(a) \(y=\dst{{c_1\over1+x}+{c_2\over1+2x}}\) (b) \(y=\dst{{c_1\over1-2x}+{c_2\over1-3x}}\) (c) \(y=\dst{{c_1\over1-2x}+{c_2x\over(1-2x)^2}}\)

(d) \(y=\dst{{c_1\over2+x}+{c_2x\over(2+x)^2}}\) (e) \(y=\dst{{c_1\over2+x}+{c_2\over2+3x}}\)

In Exercises 32–38 find the coefficients \(a_0\), …, \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

C \(y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

Show answer

\(y=\dst{1-2x-{3\over2}x^2+{5\over3}x^3+{17\over24}x^4-{11\over20}x^5+\cdots}\)
C \(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

Show answer

\(y=\dst{1-2x-{5\over2}x^2+{2\over3}x^3-{3\over8}x^4+{1\over3}x^5+\cdots}\)
C \(y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2\)

Show answer

\(y=\dst{6-2x+9x^2+{2\over3}x^3-{23\over4}x^4-{3\over10}x^5+\cdots}\)
C \(y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5\)

Show answer

\(y=\dst{2-5x+2x^2-{10\over3}x^3+{3\over2}x^4-{25\over12}x^5+\cdots}\)
C \(y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6\)

Show answer

\(y=\dst{3+6x-3x^2+x^3-2x^4-{17\over20}x^5+\cdots}\)
C \(2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2\)

Show answer

\(y=\dst{3-2x-3x^2+{3\over2}x^3+{3\over2}x^4-{49\over80}x^5+\cdots}\)
C \(3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3\)

Show answer

\(y=\dst{-2+3x+{4\over3}x^2-x^3-{19\over54}x^4+{13\over60}x^5+\cdots}\)
Find power series in \(x\) for the solutions \(y_1\) and \(y_2\) of

\[ y''+4xy'+(2+4x^2)y=0 \]

such that \(y_1(0)=1\), \(y'_1(0)=0\), \(y_2(0)=0\), \(y'_2(0)=1\), and identify \(y_1\) and \(y_2\) in terms of familiar elementary functions.

Show answer

\(\dst{y_1=\sum^\infty_{m=0} {(-1)^mx^{2m}\over m!}=e^{-x^2}, \quad y_2=\sum^\infty_{m=0} {(-1)^mx^{2m+1}\over m!}=xe^{-x^2}}\)

In Exercises 40–49 find the coefficients \(a_0\), …, \(a_N\) for \(N\) at least \(7\) in the series solution

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

of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

C \((1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3\)

Show answer

\(y=\dst{-2+3x+x^2-{1\over6}x^3-{3\over4}x^4+{31\over120}x^5+\cdots}\)
C \(y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3\)

Show answer

\(y=\dst{2+3x-{7\over2}x^2-{5\over6}x^3+{41\over24}x^4+{41\over120}x^5+\cdots}\)
C \((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)

Show answer

\(y=\dst{-3+5x-5x^2+{23\over6}x^3-{23\over12}x^4+{11\over30}x^5+\cdots}\)
C \((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)

Show answer

\(y=\dst{-2+3(x-1)+{3\over2}(x-1)^2-{17\over12}(x-1)^3-{1\over12}(x-1)^4+ {1\over8}(x-1)^5+\cdots}\)
C \(y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3\)

Show answer

\(y=\dst{2-3(x+2)+{1\over2}(x+2)^2-{1\over3}(x+2)^3+{31\over24}(x+2)^4- {53\over120}(x+2)^5+\cdots}\)
C \((1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2\)

Show answer

\(y=\dst{1-2x+{3\over2}x^2-{11\over6}x^3+{15\over8}x^4-{71\over60}x^5+\cdots}\)
C \((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

Show answer

\(y=\dst{2-(x+2)-{7\over2}(x+2)^2-{43\over6}(x+2)^3-{203\over24}(x+2)^4-{167 \over30}(x+2)^5+\cdots}\)
C \((1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1\)

Show answer

\(y=\dst{2-x-x^2+{7\over6}x^3-x^4+{89\over120}x^5+\cdots}\)
C \((x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0\)

Show answer

\(y=\dst{1+{3\over2}(x-1)^2+{1\over6}(x-1)^3-{1\over8}(x-1)^5+\cdots}\)
C \((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2\)

Show answer

\(y=\dst{1-2(x-3)+{1\over2}(x-3)^2-{1\over6}(x-3)^3+{1\over4}(x-3)^4 -{1\over6}(x-3)^5+\cdots}\)