3.3E: Exercises

In Exercises \((3.3E.1)\) to \((3.3E.12)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

Exercise \(\PageIndex{1}\)

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

Exercise \(\PageIndex{2}\)

\((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((1+x+2x^2)y''+(2+8x)y'+4y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
+2\sum_{n=2}^\infty n(n-1)a_nx^n
+2\sum_{n=1}^\infty na_nx^{n-1}
+8\sum_{n=1}^\infty na_nx^n
+4\sum_{n=0}^\infty a_nx^n
=\sum_{n=0}^\infty(n+2)(n+1)(a_{n+2}+a_{n+1}+2a_n)x^n=0\)
if
\(a_{n+2}=-a_{n+1}-2a_n, a_n\ge0\).
Starting with \(a_0=-1\) and \(a_1=2\) yields
\(y=\displaystyle{-1+2x-4x^3+4x^4+4x^5-12x^6+4x^7+\cdots}\).

Exercise \(\PageIndex{3}\)

\((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)

Exercise \(\PageIndex{4}\)

\((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((1+x+3x^2)y''+(2+15x)y'+12y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
+3\sum_{n=2}^\infty n(n-1)a_nx^n
+2\sum_{n=1}^\infty na_nx^{n-1}
+15\sum_{n=1}^\infty na_nx^n
+12\sum_{n=0}^\infty a_nx^n
=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)(n+2)a_{n+1}+3(n+2)^2a_n]x^n=0\)
if
\(a_{n+2}=-a_{n+1}-\displaystyle{3(n+2)\over n+1}a_n\),
\(a_n\ge0\). Starting with \(a_0=0\) and \(a_1=1\) yields
\(y=\displaystyle{x-x^2-{7\over2}x^3+{15\over2}x^4+{45\over8}x^5
-{261\over8}x^6+{207\over16}x^7+\cdots}\).

Exercise \(\PageIndex{5}\)

\((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)

Exercise \(\PageIndex{6}\)

\((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((3+3x+x^2)y''+(6+4x)y'+2y=
3\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+3\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
+\sum_{n=2}^\infty n(n-1)a_nx^n
+6\sum_{n=1}^\infty na_nx^{n-1}
+4\sum_{n=1}^\infty na_nx^n
+2\sum_{n=0}^\infty a_nx^n
=\sum_{n=0}^\infty(n+2)(n+1)[3a_{n+2}+3a_{n+1}+a_n]x^n=0\)
if
\(a_{n+2}=-a_{n+1}-a_n/3\),
\(a_n\ge0\). Starting with \(a_0=7\) and \(a_1=3\) yields
\(y=\displaystyle{7+3x-{16\over3}x^2+{13\over3}x^3-{23\over9}x^4+{10\over9}x^5
-{7\over27}x^6-{1\over9}x^7+\cdots}\).

Exercise \(\PageIndex{7}\)

\((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)

Exercise \(\PageIndex{8}\)

\((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)

Answer

 The equation is equivalent to
 \((1+t+2t^2)y''+(2+6t)y'+2y=0\) with \(t=x-1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+t+2t^2)y''+(2+6t)y'+2y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+2\sum_{n=2}^\infty n(n-1)a_nt^n
+2\sum_{n=1}^\infty na_nt^{n-1}
+6\sum_{n=1}^\infty na_nt^n
+2\sum_{n=0}^\infty a_nt^n
=\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}+(n+2)a_{n+1}+2(n+1)a_n]t^n=0\)
Thus,
\(a_{n+2}=-\displaystyle a_{n+1}-{2(n+1)\over n+2}a_n\),
\(a_n\ge0\). Starting with \(a_0=1\) and \(a_1=-1\) yields
\(y=\displaystyle{1-(x-1)+{4\over3}(x-1)^3-{4\over3}(x-1)^4-{4\over5}(x-1)^5
+{136\over45}(x-1)^6-{104\over63}(x-1)^7+\cdots}\)

Exercise \(\PageIndex{9}\)

\((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)

Exercise \(\PageIndex{10}\)

\((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)

Answer

The equation is equivalent to
\((1+t+t^2)y''+(3+4t)y'+2y=0\) with \)t=x-1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+t+t^2)y''+(3+4t)y'+2y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+\sum_{n=2}^\infty n(n-1)a_nt^n
+3\sum_{n=1}^\infty na_nt^{n-1}
+4\sum_{n=1}^\infty na_nt^n
+2\sum_{n=0}^\infty a_nt^n
=\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}+(n+3)a_{n+1}+(n+2)a_n]t^n=0\)
if
\(a_{n+2}=-\displaystyle{n+3\over n+2}a_{n+1}-a_n\),
\(a_n\ge0\). Starting with \(a_0=2\) and \(a_1=-1\) yields
\(y=\displaystyle{2-(x-1)-{1\over2}(x-1)^2+{5\over3}(x-1)^3-{19\over12}(x-1)^4
+{7\over30}(x-1)^5+{59\over45}(x-1)^6-{1091\over630}(x-1)^7+\cdots}\)

Exercise \(\PageIndex{11}\)

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

Exercise \(\PageIndex{12}\)

\(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)

Answer

 The equation is equivalent to
 \((1+2t+t^2)y''+(1+7t)y'+8y=0\) with \(t=x-1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+2t+t^2)y''+(1+7t)y'+8y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+\sum_{n=2}^\infty n(n-1)a_nt^n
+\sum_{n=1}^\infty na_nt^{n-1}
+7\sum_{n=1}^\infty na_nt^n
+8\sum_{n=0}^\infty a_nt^n
=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)(2n+1)a_{n+1}+(n+2)(n+4)a_n]t^n=0\)
if
\(a_{n+2}=-\displaystyle{2n+1\over n+2}a_{n+1}-{n+4\over n+1}a_n\),
\(a_n\ge0\). Starting with \(a_0=1\) and \(a_1=-2\) yields
\(y=\displaystyle{1-2(x-1)-3(x-1)^2+8(x-1)^3-4(x-1)^4-{42\over5}(x-1)^5
+19(x-1)^6-{604\over35}(x-1)^7+\cdots}\)

Exercise \(\PageIndex{13}\)

Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).

(a) Use differential equations software to solve the initial value problem

\begin{equation}\label{eq:3.3E.1}
(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}

numerically on \((-r,r)\). (See Example \((3.3.1)\).)

(b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.3E.1}, and graph

\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}

and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.

Exercise \(\PageIndex{14}\)

Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<2\).

(a) Use differential equations software to solve the initial value problem

\begin{equation}\label{eq:3.3E.2}
(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1,
\end{equation}

numerically on \((-1-r,-1+r)\). (See Example \((3.3.2)\). Why this interval?)

(b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2,\dots,a_N\) in the power series solution

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

of \eqref{eq:3.3E.2}, and graph

\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_n(x+1)^n
\end{eqnarray*}

and the solution obtained in (a) on \((-1-r,-1+r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.

Exercise \(\PageIndex{15}\)

Do the following experiment for several choices of \(a_0\), \(a_1\), and \(r\), with \(r>0\).

(a) Use differential equations software to solve the initial value problem

\begin{equation}\label{eq:3.3E.3}
y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}

numerically on \((-r,r)\). (See Example \((3.3.3)\).)

(b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.3E.3}, and graph

\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}

and the solution obtained in (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs.

Exercise \(\PageIndex{16}\)

Do the following experiment for several choices of \(a_0\) and \(a_1\).

(a) Use differential equations software to solve the initial value problem

\begin{equation}\label{eq:3.3E.4}
(1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}

numerically on \((-r,r)\).

(b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^Na_nx^n\) of \eqref{eq:3.3E.4}, and graph

\begin{eqnarray*}
T_N(x)=\sum_{n=0}^N a_nx^n
\end{eqnarray*}

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 \(r\to1\)?

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((1-x)y''-(2-x)y'+y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
-\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
-2\sum_{n=1}^\infty na_nx^{n-1}
+\sum_{n=1}^\infty na_nx^n
+\sum_{n=0}^\infty a_nx^n
=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-(n+2)(n+1)a_{n+1}+(n+1)a_n]x^n=0\)
if
\(a_{n+2}=a_{n+1}-\displaystyle{a_n\over n+2}, a_n\ge0\).

Exercise \(\PageIndex{17}\)

Follow the directions of Exercise \((3.3E.16)\) for the initial value problem

\begin{eqnarray*}
(1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
\end{eqnarray*}

Exercise \(\PageIndex{18}\)

Follow the directions of Exercise \((3.3E.16)\) for the initial value problem

\begin{eqnarray*}
(1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
\end{eqnarray*}

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((1+x^2)y''+y'+2y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nx^n
+\sum_{n=1}^\infty na_nx^{n-1}
+2\sum_{n=0}^\infty a_nx^n
=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)a_{n+1}+(n^2-n+2)a_n]x^n=0\)
if
 \(a_{n+2}=-\displaystyle{1\over n+2}a_{n+1}-
\displaystyle{n^2-n+2\over(n+2)(n+1)}a_n\).

In Exercises \((3.3E.19)\) to \((3.3E.28)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution

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

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

Exercise \(\PageIndex{19}\)

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

Exercise \(\PageIndex{20}\)

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

Answer

 The equation is equivalent to
\((3+2t)y''+(1+2t)y'-(1-2t)y=0\) with \(t=x-1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((3+2t)y''+(1+2t)y'-(1-2t)y=
3\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+\sum_{n=1}^\infty na_nt^{n-1}
+2\sum_{n=1}^\infty na_nt^n
-\sum_{n=0}^\infty a_nt^n
+2\sum_{n=0}^\infty a_nt^{n+1}
=(6a_2+a_1-a_0)+
\sum_{n=1}^\infty[3(n+2)(n+1)a_{n+2}+(n+1)(2n+1)a_{n+1}+
(2n-1)a_n+2a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{a_1-a_0\over6}\) and
\(a_{n+2}=-\displaystyle{2n+1\over3(n+2)}a_{n+1}-{2n-1\over3(n+2)(n+1)}a_n
-{2\over3(n+2)(n+1)}a_{n-1},
n\ge1\). Starting with \(a_0=1\) and \(a_1=-2\) yields
\(y=\displaystyle{1-2(x-1)+{1\over2}(x-1)^2-{1\over6}(x-1)^3+{5\over36}(x-1)^4
-{73\over1080}(x-1)^5+\cdots}\).

Exercise \(\PageIndex{21}\)

\((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

Exercise \(\PageIndex{22}\)

\((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)

Answer

The equation is equivalent to
 \((1+t)y''+(2-2t)y'+(3+t)y=0\) with \(t=x+3\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+t)y''+(2-2t)y'+(3+t)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+2\sum_{n=1}^\infty na_nt^{n-1}
-2\sum_{n=1}^\infty na_nt^n
+3\sum_{n=0}^\infty a_nt^n
+\sum_{n=0}^\infty a_nt^{n+1}
=(2a_2+2a_1+3a_0)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(n+2)(n+1)a_{n+1}
-(2n-3)a_n+a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{2a_1+3a_0\over2}\) and
\(a_{n+2}=\displaystyle-a_{n+1}+{(2n-3)a_n-a_{n-1}\over(n+2)(n+1)},
n\ge1\). Starting with \(a_0=2\) and \(a_1=-2\) yields

\(y=\displaystyle{2-2(x+3)-(x+3)^2+(x+3)^3-{11\over12}(x+3)^4+
{67\over60}(x+3)^5+\cdots}\).

Exercise \(\PageIndex{23}\)

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

Exercise \(\PageIndex{24}\)

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

Answer

The equation is equivalent to
 \((1+2t)y''+3y'+(1-t)y=0\) with \(t=x+1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+2t)y''+3y'+(1-t)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+3\sum_{n=1}^\infty na_nt^{n-1}
+\sum_{n=0}^\infty a_nt^n
-\sum_{n=0}^\infty a_nt^{n+1}
=(2a_2+3a_1+a_0)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(2n+3)(n+1)a_{n+1}
+a_n-a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{3a_1+a_0\over2}\) and
\(a_{n+2}=\displaystyle-{2n+3\over n+2}a_{n+1}-{a_n-a_{n-1}\over(n+2)(n+1)},
n \ge 1\). Starting with \(a_0=2\) and \(a_1=-3\) yields
\(y=\displaystyle{2-3(x+1)+{7\over2}(x+1)^2-5(x+1)^3+{197\over24}(x+1)^4
-{287\over20}(x+1)^5+\cdots}\).

Exercise \(\PageIndex{25}\)

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

Exercise \(\PageIndex{26}\)

\((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)

Answer

 The equation is equivalent to
 \((6-2t)y''+(3+t)y=0\) with \(t=x-2\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((6-2t)y''+(3+t)y=
6\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
-2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+3\sum_{n=0}^\infty a_nt^n
+\sum_{n=0}^\infty a_nt^{n+1}
=(12a_2+3a_0)+
\sum_{n=1}^\infty[6(n+2)(n+1)a_{n+2}-2(n+1)na_{n+1}+3a_n
+a_{n-1}]t^n=0\)  if \)a_2=-\displaystyle{a_0\over4}\) and
\(a_{n+2}=\displaystyle{n\over3(n+2)}a_{n+1}-{3a_n+a_{n-1}\over6(n+2)(n+1)},
n\ge1\). Starting with \(a_0=2\) and \(a_1=-4\) yields
\(y=\displaystyle{2-4(x-2)-{1\over2}(x-2)^2+{2\over9}(x-2)^3+{49\over432}(x-2)^4
+{23\over1080}(x-2)^5+\cdots}\).

Exercise \(\PageIndex{27}\)

\((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)

Exercise \(\PageIndex{28}\)

\((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)

Answer

 The equation is equivalent to
 \((2+4t)y''-(1-2t)y=0\) with \(t=x+4\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((2+4t)y''-(1-2t)y=
2\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+4\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
-\sum_{n=0}^\infty a_nt^n
+2\sum_{n=0}^\infty a_nt^{n+1}
=(4a_2-a_0)+
\sum_{n=1}^\infty[2(n+2)(n+1)a_{n+2}+4(n+1)na_{n+1}
-a_n+2a_{n-1}]t^n=0\)  if \)a_2=\displaystyle{a_0\over4}\) and
\(a_{n+2}=\displaystyle-{2n\over
n+2}a_{n+1}+{a_n-2a_{n-1}\over2(n+2)(n+1)},
n\ge1\). Starting with \(a_0=-1\) and \(a_1=2\) yields
\(y=\displaystyle{-1+2(x+1)-{1\over4}(x+1)^2+{1\over2}(x+1)^3-{65\over96}(x+1)^4
+{67\over80}(x+1)^5+\cdots}\).

N=5; b=zeros(N,1); b(1)=-1;b(2)=2; b(3)=b(1)/4;
for n=1:N-2
   b(n+3)=-2*n*b(n+2)/(n+2)+(b(n+1)-2*b(n))/(2*(n+2)*(n+1));
end
 

Exercise \(\PageIndex{29}\)

Show that the coefficients in the power series in \(x\) for the general solution of

\begin{eqnarray*}
(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0
\end{eqnarray*}

satisfy the recurrence relation

\begin{eqnarray*}
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.
\end{eqnarray*}

Answer

Let \(Ly=(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\(Ly=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+
\alpha\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+
\beta\sum_{n=2}^\infty n(n-1)a_nx^n+
\gamma\sum_{n=1}^\infty na_nx^{n-1}+
\delta\sum_{n=1}^\infty na_nx^n+
\epsilon\sum_{n=0}^\infty a_nx^n=
\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n+
\alpha\sum_{n=0}^\infty (n+1)na_{n+1}x^n+
\beta\sum_{n=0}^\infty n(n-1)a_nx^n+
\gamma\sum_{n=0}^\infty (n+1)a_{n+1}x^n+
\delta\sum_{n=0}^\infty na_nx^n+
\epsilon\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_nx^n\),
where
\(b_n=(n+1)(n+2)a_{n+2}+(n+1)(\alpha
n+\gamma)a_{n+1}+[\beta n(n-1)+\delta n+\epsilon]a_n\), which implies
the conclusion.

Exercise \(\PageIndex{30}\)

(a) Let \(\alpha\) and \(\beta\) be constants, with \(\beta\ne0\). Show that \(y=\sum_{n=0}^\infty a_nx^n\) is a solution of

\begin{equation}\label{eq:3.3E.5}
(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0
\end{equation}

if and only if

\begin{equation}\label{eq:3.3E.6}
a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0.
\end{equation}

An equation of this form is called a \( \textcolor{blue}{\mbox{second order homogeneous linear difference equation}} \). The polynomial \(p(r)=r^2+\alpha r+\beta\) is called the \( \textcolor{blue}{\mbox{characteristic polynomial}} \) of \eqref{eq:3.3E.6}. If \(r_1\) and \(r_2\) are the zeros of \(p\), then \(1/r_1\) and \(1/r_2\) are the zeros of

\begin{eqnarray*}
P_0(x)=1+\alpha x+\beta x^2.
\end{eqnarray*}

(b) 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

\begin{eqnarray*}
a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0
\end{eqnarray*}

satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form

\begin{eqnarray*}
y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n
\end{eqnarray*}

is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

(c) Use (b) and the formula for the sum of a geometric series to show that the functions

\begin{eqnarray*}
y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}
\end{eqnarray*}

form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

(d) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that doesn't contain either \(1/r_1\) or \(1/r_2\).

(e) 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

\begin{eqnarray*}
a_n=(c_1+c_2n)r_1^n,\quad n\ge0
\end{eqnarray*}

satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form

\begin{eqnarray*}
y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n
\end{eqnarray*}

is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

(f) Use (e) and the formula for the sum of a geometric series to show that the functions

\begin{eqnarray*}
y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}
\end{eqnarray*}

form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

(g) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that does not contain \(1/r_1\).

Answer

\part{a} Let \(\gamma=2\alpha, \delta=4\beta\), and \(\epsilon=2\beta\)
in Exercise~3.3.29 to obtain (B).

\part{b} If \(a_n=c_1r_1^n+c_2r_2^n\), then \(a_{n+2}+\alpha
a_{n+1}+\beta a_n=c_1r_1^n(r_1^2+\alpha
r+\beta)+c_2r_2^n(r_2^2+\alpha
r_2+\beta)=c_1r_1^nP_0(r_1)+c_2r_2^nP_0(r_2)=0\), so \(\{a_n\}\)
satisfies (B). Since \(1/r_1\) and \(1/r_2\) are the zeros
of \)P_0\), Theorem~3.2.1 implies  that \(\sum_{n=0}^\infty
(c_1r_1^n+c_2r_2^n)x^n\) is a solution of (A) on \((-\rho,\rho)\).

\part{c} If \(|x|<\rho\), then \(|r_1x|<\rho\) and \(|r_2x|<1\), so
\(\displaystyle\sum_{n=0}^\infty r_i^nx^n={1\over1-r_ix}=y_i, i=1,2\).
Therefore,
\part{b} implies that \(\{y_1,y_2\}\) is a fundamental set of solutions
of (A) on \((-\rho,\rho)\).

\part{d} (A) can written as \(P_0y''+2P_0'y'+P_0''y=(P_0y)''=0\).
Therefore, \(P_0y=a+bx\) where \(a\) and \(b\) are arbitrary constants, and
 a partial fraction expansion shows that  the general solution of
(A)
on any interval not containing \(1/r_1\) or \(1/r_2\) is \(y=\displaystyle{a+bx\over
P_0(x)}={c_1\over1-r_1x}+{c_2\over1-r_2x}=c_1y_1+c_2y_2\).

\part{e} If \(a_n=c_1r_1^n+c_2r_2^n\), then \(a_{n+2}+\alpha
a_{n+1}+\beta a_n=c_1r_1^n(r_1^2+\alpha
r+\beta)+c_2r_1^n[(n+2)r_1^2+\alpha(n+1)
r_2+\beta n)=(c_1+nc_2)r_1^nP_0(r_1)+c_2r_1^nP_0'(r_1)=0\), so
\)\{a_n\}\) satisfies (B). Since \(1/r_1\) is the only  zero
of \(P_0\), Theorem~3.2.1 implies  that \(\sum_{n=0}^\infty
(c_1+c_2n)r_1^n)x^n\) is a solution of (A) on \((-\rho,\rho)\).

\part{f} If \(|x|<\rho\), then \(|r_1x|<\rho\), so
\(\displaystyle\sum_{n=0}^\infty r_1^nx^n={1\over1-r_1x}=y_1\).
Differentiating this and multiplying the result by \(x\)
shows that
\(\displaystyle\sum_{n=0}^\infty nr_1^nx^n={r_1x\over(1-r_1x)^2}=r_1y_2\).
Therefore,
\part{e} implies that \(\{y_1,y_2\}\) is a fundamental set of solutions
of (A) on \((-\rho,\rho)\).

\part{g} The argument is the same as in \part{c}, but now the
partial fraction expansion can be written as
 \(y=\displaystyle{a+bx\over
P_0(x)}={c_1\over1-r_1x}+{c_2x\over(1-r_2x)^2}=c_1y_1+c_2y_2\).

Exercise \(\PageIndex{31}\)

Use the results of Exercise \((3.3E.30)\) 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) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

In Exercises \((3.3E.32)\) to \((3.3E.38)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

Exercise \(\PageIndex{32}\)

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

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\(y''+2xy'+(3+2x^2)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+2\sum_{n=1}^\infty na_nx^n
+3\sum_{n=0}^\infty a_nx^n
+2\sum_{n=0}^\infty a_nx^{n+2}
=(2a_2+3a_0)+(6a_3+5a_1)x+
\sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}+(2n+3)a_n+2a_{n-2}]x^n=0\) if
\(a_2=-3a_0/2, a_3=-5a_1/6\), and
\(a_{n+2}=-\displaystyle{(2n+3)a_n+2a_{n-2}\over(n+2)(n+1)},
 n\ge2\).
Starting with \(a_0=1\) and \(a_1=-2\) yields
\(y=\displaystyle{1-2x-{3\over2}x^2+{5\over3}x^3+{17\over24}x^4-{11\over20}x^5+\cdots}\).

Exercise \(\PageIndex{33}\)

\(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)

Exercise \(\PageIndex{34}\)

\(y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\(y''+5xy'-(3-x^2)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+5\sum_{n=1}^\infty na_nx^n
-3\sum_{n=0}^\infty a_nx^n
+\sum_{n=0}^\infty a_nx^{n+2}
=(2a_2-3a_0)+(6a_3+2a_1)x+
\sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}+(5n-3)a_n+a_{n-2}]x^n=0\) if
\(a_2=3a_0/2, a_3=-a_1/3\), and
\(a_{n+2}=-\displaystyle{(5n-3)a_n+a_{n-2}\over(n+2)(n+1)},
 n\ge2\).
Starting with \(a_0=6\) and \(a_1=-2\) yields
\(y=\displaystyle{6-2x+9x^2+{2\over3}x^3-{23\over4}x^4-{3\over10}x^5+\cdots}\).

Exercise \(\PageIndex{35}\)

\(y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5\)

Exercise \(\PageIndex{36}\)

\(y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\(y''-3xy'+(2+4x^2)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
-3\sum_{n=1}^\infty na_nx^n
+2\sum_{n=0}^\infty a_nx^n
+4\sum_{n=0}^\infty a_nx^{n+2}
=(2a_2+2a_0)+(6a_3-a_1)x+
\sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}-(3n-2)a_n+4a_{n-2}]x^n=0\) if
\(a_2=-a_0, a_3=a_1/6\), and
\(a_{n+2}=\displaystyle{(3n-2)a_n-4a_{n-2}\over(n+2)(n+1)},
 n\ge2\).
Starting with \(a_0=3\) and \(a_1=6\) yields
\(y=\displaystyle{3+6x-3x^2+x^3-2x^4-{17\over20}x^5+\cdots}\).

Exercise \(\PageIndex{37}\)

\(2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2\)

Exercise \(\PageIndex{38}\)

\(3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3\)

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\(3y''+2xy'+(4-x^2)y=
3\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+2\sum_{n=1}^\infty na_nx^n
+4\sum_{n=0}^\infty a_nx^n
-\sum_{n=0}^\infty a_nx^{n+2}
=(6a_2+4a_0)+(18a_3+6a_1)x+
\sum_{n=2}^\infty[3(n+2)(n+1)a_{n+2}+(2n+4)a_n-a_{n-2}]x^n=0\) if
\(a_2=-2a_0/3, a_3=-a_1/3\), and
\(a_{n+2}=-\displaystyle{(2n+4)a_n-a_{n-2}\over3(n+2)(n+1)},
 n\ge2\).
Starting with \(a_0=-2\) and \(a_1=3\) yields
\(y=\displaystyle-2+3x+{4\over3}x^2-x^3-{19\over54}x^4+{13\over60}x^5+\cdots\).

Exercise \(\PageIndex{39}\)

Find power series in \(x\) for the solutions \(y_1\) and \(y_2\) of

\begin{eqnarray*}
y''+4xy'+(2+4x^2)y=0
\end{eqnarray*}

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.

 

 

In Exercises \((3.3E.40)\) tp \((3.3E.49)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution

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

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

Exercise \(\PageIndex{40}\)

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

Answer

If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\((1+x)y''+x^2y'+(1+2x)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+ \sum_{n=2}^\infty n(n-1)a_nx^{n-1}
+\sum_{n=1}^\infty na_nx^{n+1}
+\sum_{n=0}^\infty a_nx^n
+2\sum_{n=0}^\infty a_nx^{n+1}
=(2a_2+a_0)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(n+1)na_{n+1}+a_n+(n+1)a_{n-1}]x^n=0\)
if
\(a_2=-a_0/2\) and
\(a_{n+2}=-\displaystyle{(n+1)na_{n+1}+a_n+(n+1)a_{n-1}\over(n+2)(n+1)},
 n \ge 1\).
Starting with \(a_0=-2\) and \(a_1=3\) yields
\(y=\displaystyle{-2+3x+x^2-{1\over6}x^3-{3\over4}x^4+{31\over120}x^5+\cdots}\).

Exercise \(\PageIndex{41}\)

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

Exercise \(\PageIndex{42}\)

\((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)

Answer

If \)y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
\)(1+x^2)y''+(2+x^2)y'+xy=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
+\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^n
+2\sum_{n=1}^\infty na_nx^{n-1}
+\sum_{n=1}^\infty na_nx^{n+1}
+\sum_{n=1}^\infty a_nx^{n+1}
=(2a_2+2a_1)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+2(n+1)a_{n+1}+n(n-1)a_n+na_{n-1}]x^n=0\)
if
\(a_2=-a_1\) and
\(a_{n+2}=-\displaystyle{[2(n+1)a_{n+1}+n(n-1)a_n+na_{n-1}]\over(n+2)(n+1)},
 n \ge 1\).
Starting with \(a_0=-3\) and \(a_1=5\) yields
\(y=\displaystyle{-3+5x-5x^2+{23\over6}x^3-{23\over12}x^4+{11\over30}x^5+\cdots}\).

Exercise \(\PageIndex{43}\)

\((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)

Exercise \(\PageIndex{44}\)

\(y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3\)

Answer

The equation is equivalent to
 \(y''+(1+3t^2)y'+(1+2t)y=0\) with \)t=x-2\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\(y''+(1+3t^2)y'+(1+2t)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+\sum_{n=1}^\infty na_nt^{n-1}
+3\sum_{n=1}^\infty na_nt^{n+1}
+\sum_{n=0}^\infty a_nt^n
+2\sum_{n=0}^\infty a_nt^{n+1}=(2a_2+a_1+a_0)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+
(n+1)a_{n+1}+a_n+(3n-1)a_{n-1}]t^n=0\) if \)a_2=-(a_1+a_0)/2\) and
\)a_{n+2}=-\displaystyle{[(n+1)a_{n+1}+a_n+(3n-1)a_{n-1}]\over
(n+2)(n+1)},
n \ge 1 \). Starting with \(a_0=2\) and \(a_1=-3\) yields

\(y=\displaystyle{2-3(x+2)+{1\over2}(x+2)^2-{1\over3}(x+2)^3+{31\over24}(x+2)^4-
{53\over120}(x+2)^5+\cdots}\).

Exercise \(\PageIndex{45}\)

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

Exercise \(\PageIndex{46}\)

\((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)

Answer

The equation is equivalent to
\((1-t^2)y''-(7-8t+t^2)y'+ty=0\) with \)t=x+2\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1-t^2)y''-(7-8t+t^2)y'+ty=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
- \sum_{n=2}^\infty n(n-1)a_nt^n
-7\sum_{n=1}^\infty na_nt^{n-1}
+8\sum_{n=1}^\infty na_nt^n
-\sum_{n=1}^\infty na_nt^{n+1}
+\sum_{n=0}^\infty a_nt^{n+1}
=(2a_2-7a_1)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}
-7(n+1)a_{n+1}-n(n-9)a_n-(n-2)a_{n-1}]t^n=0\) if \)a_2=7a_1/2\)
and
\(a_{n+2}=\displaystyle{[7(n+1)a_{n+1}+n(n-9)a_n+(n-2)a_{n-1}]\over
(n+2)(n+1)},
n\ge1\). Starting with \(a_0=2\) and \(a_1=-1\) yields

\(y=\displaystyle{2-(x+2)-{7\over2}(x+2)^2-{43\over6}(x+2)^3-{203\over24}(x+2)^4-{167
\over30}(x+2)^5+\cdots}\).

Exercise \(\PageIndex{47}\)

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

Exercise \(\PageIndex{48}\)

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

Answer

The equation is equivalent to
 \((1+3t+2t^2)y''-(3+t-t^2)y'-(3+t)y=0\) with \(t=x-1\).
If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
\((1+3t+2t^2)y''-(3+t-t^2)y'-(3+t)y=
\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
+3\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
+2\sum_{n=2}^\infty n(n-1)a_nt^n
-3\sum_{n=1}^\infty na_nt^{n-1}
-\sum_{n=1}^\infty na_nt^n
+\sum_{n=1}^\infty na_nt^{n+1}
-3\sum_{n=0}^\infty a_nt^n
-\sum_{n=0}^\infty a_nt^{n+1}
=(2a_2-3a_1-3a_0)+
\sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}
+3(n^2-1)a_{n+1}+(2n^2-3n-3)(n+1)a_n+(n-2)a_{n-1}]t^n=0\) if
\(a_2=3(a_1+a_0)/2\) and
\(a_{n+2}=-\displaystyle{[3(n^2-1)a_{n+1}+(2n^2-3n-3)(n+1)a_n+(n-2)a_{n-1}]\over
(n+2)(n+1)},
n\ge1\). Starting with \(a_0=1\) and \(a_1=0\) yields
\(y=\displaystyle{1+{3\over2}(x-1)^2+{1\over6}(x-1)^3-{1\over8}(x-1)^5+\cdots}\).

Exercise \(\PageIndex{49}\)

\((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2\)