3.2E: Exercises

Exercises \((3.2E.1)\) to \((3.2E.8)\), find the power series in \(x\) for the general solution.

Exercise \(\PageIndex{1}\)

\((1+x^2)y''+6xy'+6y=0\)

Exercise \(\PageIndex{2}\)

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

Answer

\(p(n)=n(n-1)+2n-2=(n+2)(n-1)\) ;
\(a_{n+2}=-{n-1\over n+1}a_n\) ;
\(a_{2m+2}=-\displaystyle {2m-1\over 2m+1}a_{2m}\) , so
\(a_{2m}=\displaystyle {(-1)^m\over 2m-1}a_0\) ;
\(a_{2m+3}=\displaystyle -{m\over m+2}a_{2m+1}=0\)
if \(m\ge0\) ;
\(y=\displaystyle {a_0\sum_{m=0}^\infty(-1)^{m+1}{x^{2m}\over 2m-1}+a_1x}\) .

Exercise \(\PageIndex{3}\)

\((1+x^2)y''-8xy'+20y=0\)

Exercise \(\PageIndex{4}\)

\((1-x^2)y''-8xy'-12y=0\)

Answer

\(p(n)=-n(n-1)-8n-12=-(n+3)(n+4)\);
\(a_{n+2}={(n+3)(n+4)\over(n+2)(n+1)}a_n\);
\(a_{2m+2}=\displaystyle{(m+2)(2m+3)\over(m+1)(2m+1)}a_{2m}\), so
\(a_{2m}=(m+1)(2m+1)a_0\);
\(a_{2m+3}=\displaystyle{(m+2)(2m+5)\over(m+1)(2m+3)}a_{2m+1}\)
 so \(a_{2m+1}=\displaystyle{(m+1)(2m+3)\over3}a_1\);
\(y=\displaystyle{a_0\sum_{m=0}^\infty(m+1)(2m+1)x^{2m}+{a_1\over3}\sum_{m=0}^\infty
(m+1)(2m+3)x^{2m+1}}\).

Exercise \(\PageIndex{5}\)

\((1+2x^2)y''+7xy'+2y=0\)

Exercise \(\PageIndex{6}\)

\(\displaystyle{(1+x^2)y''+2xy'+{1\over4}y=0}\)

Answer

\( p(n) = n(n-1) + 2n + \frac{1}{4} = \frac{(2n+1)^2}{4} \) ,\( a_{n+2} = -\frac{(2n+1)^2}{4(n+2)(n+1)} a_n \) \( a_{2m+2} = -\frac{(4m+1)^2}{8(m+1)(2m+1)} a_{2m} \), \( a_{2m} = (-1)^m \left[ \prod_{j=0}^{m-1} \frac{(4j+1)^2}{2j+1} \right] \frac{1}{8^m m!} a_0 \) ,\( a_{2m+3} = -\frac{(4m+3)^2}{8(2m+3)(m+1)} a_{2m+1} \) ,\( a_{2m+1} = (-1)^m \left[ \prod_{j=0}^{m-1} \frac{(4j+3)^2}{2j+3} \right] \frac{1}{8^m m!} a_1 \) ,\( y = a_0 \sum_{m=0}^\infty (-1)^m \left[ \prod_{j=0}^{m-1} \frac{(4j+1)^2}{2j+1} \right] \frac{x^{2m}}{8^m m!} + a_1 \sum_{m=0}^\infty (-1)^m \left[ \prod_{j=0}^{m-1} \frac{(4j+3)^2}{2j+3} \right] \frac{x^{2m+1}}{8^m m!} \)

Exercise \(\PageIndex{7}\)

\((1-x^2)y''-5xy'-4y=0\)

Exercise \(\PageIndex{8}\)

\((1+x^2)y''-10xy'+28y=0\)

Answer

 

\(
p(n) = n(n-1) - 10n + 28 = (n-7)(n-4)
\)

\(
a_{n+2} = -\frac{(n-7)(n-4)}{(n+2)(n+1)} a_n
\)

\(
a_{2m+2} = -\frac{2(2m-7)(m-2)}{2(m+1)(2m+1)} a_{2m}
\)

\(
a_2 = -14 a_0, \quad a_4 = -\frac{5}{6} a_2 = \frac{35}{3} a_0
\)

\(
a_{2m} = 0 \quad \text{if } m \geq 3
\)

\(
a_{2m+3} = -\frac{(m-3)(2m-3)}{(2m+3)(m+1)} a_{2m+1}
\)

\(
a_3 = -3 a_1, \quad a_5 = -\frac{1}{5} a_3 = \frac{3}{5} a_1
\)

\(
a_7 = \frac{1}{21} a_5 = \frac{1}{35} a_1
\)

\(
y = a_0 \left( 1 - 14x^2 + \frac{35}{3} x^4 \right) + a_1 \left( x - 3x^3 + \frac{3}{5} x^5 + \frac{1}{35} x^7 \right)
\)

 

Exercise \(\PageIndex{9}\)

(a) Find the power series in \(x\) for the general solution of \(y''+xy'+2y=0\).

(b) For several choices of \(a_0\) and \(a_1\), use differential equations software to solve the initial value problem

\begin{equation}\label{eq:3.2E.1}
y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}

numerically on \((-5,5)\).

(c) For fixed \(r\) in \(\{1,2,3,4,5\}\) graph

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

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

Exercise \(\PageIndex{10}\)

Follow the directions of Exercise \((3.2E.9)\) for the differential equation

\begin{eqnarray*}
y''+2xy'+3y=0.
\end{eqnarray*}

Answer

\(
p(n) = 2n + 3
\)

\(
a_{n+2} = -\frac{2n+3}{(n+2)(n+1)} a_n
\)

\(
a_{2m+2} = -\frac{4m+3}{2(m+1)(2m+1)} a_{2m}
\)

\(
a_{2m} = \left[ \prod_{j=0}^{m-1} \frac{4j+3}{2j+1} \right] \frac{(-1)^m}{2^m m!} a_0
\)

\(
a_{2m+3} = -\frac{4m+5}{2(2m+3)(m+1)} a_{2m+1}
\)

\(
a_{2m+1} = \left[ \prod_{j=0}^{m-1} \frac{4j+5}{2j+3} \right] \frac{(-1)^m}{2^m m!} a_1
\)

\(
y = a_0 \sum_{m=0}^\infty (-1)^m \left[ \prod_{j=0}^{m-1} \frac{4j+3}{2j+1} \right] \frac{x^{2m}}{2^m m!} + a_1 \sum_{m=0}^\infty (-1)^m \left[ \prod_{j=0}^{m-1} \frac{4j+5}{2j+3} \right] \frac{x^{2m+1}}{2^m m!}
\)

 

In Exercises \((3.2E.11)\) to \((3.2E.13)\), find \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

Exercise \(\PageIndex{11}\)

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

Exercise \(\PageIndex{12}\)

\((1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1\)

Answer

$$p(n)=2n(n-1)-9n-6=(n-6)(2n+1)$$;
$$a_{n+2}=\displaystyle-{(n-6)(2n+1)\over(n+2)(n+1)}a_n$$;
$$a_0=y(0)=1$$; $$a_1=y'(0)=-1$$.

Exercise \(\PageIndex{13}\)

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

Answer

$$p(n)=8n(n-1)+2=2(2n-1)^2$$;
$$a_{n+2}=\displaystyle-{2(2n-1)^2\over(n+2)(n+1)}a_n$$;
$$a_0=y(0)=2$$; $$a_1=y'(0)=-1$$.

Exercise \(\PageIndex{14}\)

Do the next 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:3.2E.2}
(1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
\end{equation}

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

(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:3.2E.2}, and graph

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

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

Exercise \(\PageIndex{15}\)

Do part (a) and part (b) for several values of \(r\) in \((0,1)\):

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

\begin{equation}\label{eq:3.2E.3}
(1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1,
\end{equation}

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

(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:3.2E.3}, and graph

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

and the solution obtained in part (a) on \((-r,r)\). Continue increasing \(N\) until there's no perceptible difference between the two graphs. What happens to the required \(N\) as \(r\to1\)?

(c) Try part (a) and part (b) with \(r=1.2\). Explain your results.

In Exercises \((3.2E.16)\) to \((3.2E.20)\), find the power series in \(x-x_0\) for the general solution.

Exercise \(\PageIndex{16}\)

\(y''-y=0;\quad x_0=3\)

Answer

$$p(n)=-1$$;
$$a_{n+2}=\displaystyle{1\over(n+2)(n+1)}a_n$$;
$$a_{2m+2}=\displaystyle{1\over(2m+2)(2m+1)}a_{2m}$$, so
$$a_{2m}=\displaystyle{1\over(2m)!}a_0$$;
$$a_{2m+3}=\displaystyle{1\over(2m+3)(2m+1)}a_{2m+1}$$,
so $$a_{2m+1}=\displaystyle{1\over(2m+1)!}a_1$$;
$$y=\displaystyle{a_0\sum_{m=0}^\infty{(x-3)^{2m}\over(2m)!}+a_1\sum_{m=0}^\infty
{(x-3)^{2m+1}\over(2m+1)!}}$$.

Exercise \(\PageIndex{17}\)

\(y''-(x-3)y'-y=0;\quad x_0=3\)

Exercise \(\PageIndex{18}\)

\((1-4x+2x^2)y''+10(x-1)y'+6y=0;\quad x_0=1\)

Answer

Let \(t=x-1\);  then $$(1-2t^2)y''-10ty'-6y=0$$;
$$p(n)=-2n(n-1)-10n-6=-2(n+1)(n+3)$$;
$$a_{n+2}=\displaystyle{2(n+3)\over n+2}a_n$$;
$$a_{2m+2}=\displaystyle{2m+3\over m+1}a_{2m}$$, so
$$a_{2m}=\displaystyle{1\over m!}\left[\prod_{j=0}^{m-1}(2j+3)\right]a_0$$;
$$a_{2m+3}=\displaystyle{4(m+2)\over 2m+3}a_{2m+1}$$,
so $$a_{2m+1}=\displaystyle{4^m(m+1)!\over\prod_{j=0}^{m-1}(2j+3)}a_1$$;
$$y=\displaystyle{a_0\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+3)\right]{
(x-1)^{2m}\over m!}}\\+\displaystyle{a_1\sum_{m=0}^\infty
{4^m(m+1)!\over\prod_{j=0}^{m-1}(2j+3)}(x-1)^{2m+1}}$$.

Exercise \(\PageIndex{19}\)

\((11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2\)

Exercise \(\PageIndex{20}\)

\((5+6x+3x^2)y''+9(x+1)y'+3y=0;\quad x_0=-1\)

Answer

Let \(t=x+1\);  then
$$\displaystyle\left(1+{3t^2\over3}\right)y''-{9t\over2}y'+{3\over2}y=0$$;
$$p(n)=\displaystyle{3\over2}n(n-1)+{9\over2}n+{3\over2}={3\over2}(n+1)^2$$;
$$a_{n+2}=-\displaystyle{3(n+1)\over2(n+2)}a_n$$;
$$a_{2m+2}=-\displaystyle{3(2m+1)\over4(m+1)}a_{2m}$$, so
$$a_{2m}=(-1)^m\displaystyle\left[\prod_{j=0}^{m-1}(2j+1)\right]{3^m\over4^mm!}a_0$$;
$$a_{2m+3}=-\displaystyle{3(m+1)\over2m+3}a_{2m+1}$$,
so $$a_{2m+1}=\displaystyle(-1)^m{3^mm!\over\prod_{j=0}^{m-1}(2j+3)}a_1$$;\\
$$y=\displaystyle{a_0\sum_{m=0}^\infty(-1)^m\left[\prod_{j=0}^{m-1}(2j+1)\right]
{3^m\over4^mm!}(x+1)^{2m}+a_1\sum_{m=0}^\infty(-1)^m{3^mm!\over
\prod_{j=0}^{m-1}(2j+3)}(x+1)^{2m+1}}$$.

In Exercises \((3.2E.21)\) to \((3.2E.26)\), find \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the power series \(y=\sum_{n=0}^\infty a_n(x-x_0)^n\) for the solution of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

Exercise \(\PageIndex{21}\)

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

Exercise \(\PageIndex{22}\)

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

Answer

$$p(n)=n+3$$;
$$a_{n+2}=-\displaystyle{n+3\over(n+2)(n+1)}a_n$$;
$$a_0=y(3)=-2$$; $$a_1=y'(3)=3$$.

Exercise \(\PageIndex{23}\)

\((5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1\)

Exercise \(\PageIndex{24}\)

\((4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6\)

Answer

Let $$t=x-3$$; $$(1+4t^2)y''+y=0$$;
$$p(n)=(4n(n-1)+1=2n-1)^2$$;
$$a_{n+2}=-\displaystyle{(2n-1)^2\over(n+2)(n+1)}a_n$$;
$$a_0=y(3)=4$$; $$a_1=y'(3)=-6$$.

Exercise \(\PageIndex{25}\)

\((x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4\)

Exercise \(\PageIndex{26}\)

\((2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3\)

Answer

Let $$t=x+1$$;
$$\displaystyle\left(1+{2t^2\over3}\right)y''-{20\over3}ty'+20y=0$$;
$$p(n)=\displaystyle{2\over3}n(n-1)-{20\over3}n+20={2(n-6)(n-5)\over3}$$;
$$a_{n+2}=-\displaystyle{2(n-6)(n-5)\over3(n+2)(n+1)}a_n$$;
$$a_0=y(-1)=3$$; $$a_1=y'(-1)=-3$$.

Exercise \(\PageIndex{27}\)

(a) Find a power series in \(x\) for the general solution of

\begin{equation}\label{eq:3.2E.4}
(1+x^2)y''+4xy'+2y=0.
\end{equation}

(b) Use (a) and the formula

\begin{eqnarray*}
{1\over1-r}=1+r+r^2+\cdots+r^n+\cdots \quad(-1<r<1)
\end{eqnarray*}

for the sum of a geometric series to find a closed form expression for the general solution of \eqref{eq:3.2E.4} on \((-1,1)\).

(c) Show that the expression obtained in part (b) is actually the general solution of \eqref{eq:3.2E.4} on \((-\infty,\infty)\).

Exercise \(\PageIndex{28}\)

Use Theorem \((3.2.2)\) to show that the power series in \(x\) for the general solution of

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

is

\begin{eqnarray*}
y=a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}
p(2j)\right] {x^{2m}\over(2m)!} +
a_1\sum^\infty_{m=0}(-1)^m
\left[\prod^{m-1}_{j=0}p(2j+1)\right]
{x^{2m+1}\over(2m+1)!}.
\end{eqnarray*}

Answer

From Theorem~3.2.2,
$$a_{n+2}=\displaystyle-{p(n) \over(n+2)(n+1)}a_n$$;
$$a_{2m+2}=\displaystyle-{p(2m)\over(2m+2)(2m+1)}a_{2m}$$, so $$a_{2m}=\displaystyle
\left[\prod^{m-1}_{j=0}p(2j)\right] {(-1)^m\over(2m)!}a_0$$;
$$a_{2m+3}=\displaystyle-{p(2m+1)\over(2m+3)(2m+2)}a_{2m}$$, so $$a_{2m+1}=\displaystyle
\left[\prod^{m-1}_{j=0}p(2j+1)\right]{(-1)^m\over(2m+1)!}a_1$$.

Exercise \(\PageIndex{29}\)

Use Exercise \((3.2E.28)\) to show that all solutions of

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

are polynomials if and only if

\begin{eqnarray*}
\alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1),
\end{eqnarray*}

where \(r\) and \(s\) are nonnegative integers.

Exercise \(\PageIndex{30}\)

(a) Use Exercise \((3.2E.28)\) to show that the power series in \(x\) for the general solution of

\begin{eqnarray*}
(1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0
\end{eqnarray*}

is \(y=a_0y_1+a_1y_2\), where

\begin{eqnarray*}
y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha)(2j+\alpha+2b-1) \right]{x^{2m}\over(2m)!}\\
\mbox{and}\\
y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha)(2j+\alpha+2b)\right]{x^{2m+1}\over(2m+1)!}.
\end{eqnarray*}

(b) Suppose \(2b\) isn't a negative odd integer and \(k\) is a nonnegative integer. Show that \(y_1\) is a polynomial of degree \(2k\) such that \(y_1(-x)=y_1(x)\) if \(\alpha=2k\), while \(y_2\) is a polynomial of degree \(2k+1\) such that \(y_2(-x)=-y_2(-x)\) if \(\alpha=2k+1\). Conclude that if \(n\) is a nonnegative integer, then there's a polynomial \(P_n\) of degree \(n\) such that \(P_n(-x)=(-1)^nP_n(x)\) and

\begin{equation}\label{eq:3.2E.5}
(1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0.
\end{equation}

(c) Show that \eqref{eq:3.2E.5} implies that

\begin{eqnarray*}
[(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n,
\end{eqnarray*}

and use this to show that if \(m\) and \(n\) are nonnegative integers, then

\begin{equation}\label{eq:3.2E.6}
\begin{array}{ll}
[(1-x^2)^bP_n']'P_m-[(1-x^2)^bP_m']'P_n=&\\
\left[m(m+2b-1)-n(n+2b-1)\right](1-x^2)^{b-1}P_mP_n.&
\end{array}
\end{equation}

(d) Now suppose \(b>0\). Use \eqref{eq:3.2E.6} and integration by parts to show that if \(m\ne n\), then

\begin{eqnarray*}
\int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0.
\end{eqnarray*}

(We say that \(P_m\) and \(P_n\) are \( \textcolor{blue}{\mbox{orthogonal on}} \) \((-1,1)\) \( \textcolor{blue}{\mbox{with respect to the weighting function}} \) \((1-x^2)^{b-1}\).)
 

Answer

a. Here
\(p(n)=-[n(n-1)+2bn-\alpha(\alpha+2b-1)]=-(n-\alpha)(n+\alpha+2b-1)\),
so Exercise~3.2.28 implies that \(y_1\) and \(y_2\) have the stated
forms. If \(\alpha=2k\), then
$$
y_1=\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-2k)(2j+2k+2b-1)
\right]{x^{2m}\over(2m)!}
\eqno{\rm(C)}.
$$
If \(\alpha=2k+1\), then
$$
y_2=\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-2k)(2j+2k+2b)
\right]{x^{2m+1}\over(2m+1)!}.
\eqno{\rm(D)}
$$
Since \(2b\) is not a negative integer and \(\prod_{j=0}^{m-1}(2j-2k)=0\)
if \(m>k, \,y_1\) in (C) and \(y_2\) in (D) have the stated properties.
This implies the conclusions regarding \(P_n\).

b. Multiplying (A) through by \((1-x^2)^{b-1}\) yields
$$
[(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n.
\eqno{\rm(E)}
$$

c.
Therefore,
$$
[(1-x^2)^b P_m']'=-m(m+2b-1)(1-x^2)^{b-1}P_m.
\eqno{\rm(F)}
$$
Subtract  \(P_n\) times (F) from \(P_m\) times (E) to obtain (B).

d. Integrating the left side of (B) by parts over \([-1,1]\)
yields zero, which implies the conclusion.
 

Exercise \(\PageIndex{31}\)

(a) Use Exercise \((3.2E.28)\) to show that the power series in \(x\) for the general solution of Hermite's equation

\begin{eqnarray*}
y''-2xy'+2\alpha y=0
\end{eqnarray*}

is \(y=a_0y_1+a_1y_1\), where

\begin{eqnarray*}
y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha) \right]{2^mx^{2m}\over(2m)!}\\
\mbox{and}\\
y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha) \right]{2^mx^{2m+1}\over(2m+1)!}.
\end{eqnarray*}

(b) Suppose \(k\) is a nonnegative integer. Show that \(y_1\) is a polynomial of degree \(2k\) such that \(y_1(-x)=y_1(x)\) if \(\alpha=2k\), while \(y_2\) is a polynomial of degree \(2k+1\) such that \(y_2(-x)=-y_2(-x)\) if \(\alpha=2k+1\). Conclude that if \(n\) is a nonnegative integer then there's a polynomial \(P_n\) of degree \(n\) such that \(P_n(-x)=(-1)^nP_n(x)\) and

\begin{equation}\label{eq:3.2E.7}
P_n''-2xP_n'+2nP_n=0.
\end{equation}

(c) Show that \eqref{eq:3.2E.7} implies that

\begin{eqnarray*}
[e^{-x^2}P_n']'=-2ne^{-x^2}P_n,
\end{eqnarray*}

and use this to show that if \(m\) and \(n\) are nonnegative integers, then

\begin{equation}\label{eq:3.2E.8}
[e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n.
\end{equation}

(d) Use \eqref{eq:3.2E.8} and integration by parts to show that if \(m\ne n\), then

\begin{eqnarray*}
\int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0.
\end{eqnarray*}

(We say that \(P_m\) and \(P_n\) are \( \textcolor{blue}{\mbox{orthogonal on}} \) \((-\infty,\infty)\) \( \textcolor{blue}{\mbox{with respect to the weighting function}} \) \(e^{-x^2}\).)

Answer

Add texts here. Do not delete this text first.

Exercise \(\PageIndex{32}\)

Consider the equation

\begin{equation}\label{eq:3.2E.9}
\left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0,
\end{equation}

and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\). (The special case \(y''-xy=0\) of \eqref{eq:3.2E.9} is Airy's equation.)

(a) Modify the argument used to prove Theorem \((3.2.2)\) to show that

\begin{eqnarray*}
y=\sum_{n=0}^\infty a_nx^n
\end{eqnarray*}

is a solution of \eqref{eq:3.2E.9} if and only if \(a_2=0\) and

\begin{eqnarray*}
a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.
\end{eqnarray*}

(b) Show from (a) that \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\), and that

\begin{eqnarray*}
a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0,\\
\mbox{and}\\
a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0,
\end{eqnarray*}

where \(a_0\) and \(a_1\) may be specified arbitrarily.

(c) Conclude from (b) that the power series in \(x\) for the general solution of \eqref{eq:3.2E.9} is

\begin{eqnarray*}
\begin{array}{l}
y=\displaystyle{a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\
\qquad\displaystyle{+a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p(3j+1)\over3j+4}\right] {x^{3m+1}\over3^mm!}}.
\end{array}
\end{eqnarray*}

Answer

a. Let $$Ly=(1+\alpha x^3)y''+\beta x^2y'+\gamma xy$$. If
$$y=\displaystyle\sum_{n=0}^\infty a_nx^n$$, then
$$Ly=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=0}^\infty
p(n)a_nx^{n+1}=
2a_2+\sum_{n=0}^\infty [(n+3)(n+2)a_{n+3}+p(n)a_n]x^{n+1}
= 0$$ if and only if $$a_2=0$$ and $$\displaystyle
a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n$$ for $$n\ge0$$.

In Exercises \((3.2E.33)\) to \((3.2E.37)\), use the method of Exercise \((3.2E.32)\) to find the power series in \(x\) for the general solution.

Exercise \(\PageIndex{33}\)

\(y''-xy=0\)

Exercise \(\PageIndex{34}\)

\((1-2x^3)y''-10x^2y'-8xy=0\)

Answer

 $$p(r)=-2r(r-1)-10r-8=-2(r+2)^2$$;
(A) $$\displaystyle\prod_{j=0}^{m-1}{p(3j)\over3j+2}=
\prod_{j=0}^{m-1}{(-2)(3j+2)^2\over3j+2}
=(-1)^m2^m\prod_{j=0}^{m-1}(3j+2)$$;
(B) $$\displaystyle\prod_{j=0}^{m-1}{p(3j+1)\over3j+4}=
\prod_{j=0}^{m-1}{(-2)(3j+3)^2\over3j+4}
={(-1)^m2^m(m!)^2\over\prod_{j=0}^{m-1}(3j+4)}$$.
Substituting (A) and (B) into the
result of Exercise~7.2.32\part{c} yields

$$y=\displaystyle{a_0\sum_{m=0}^\infty
\left(2\over3\right)^m\left[\prod_{j=0}^{m-1}(3j+2)\right]{x^{3m}\over m!}
+a_1\sum_{m=0}^\infty{6^mm!\over\prod_{j=0}^{m-1}(3j+4)}x^{3m+1}}$$.

Exercise \(\PageIndex{35}\)

\((1+x^3)y''+7x^2y'+9xy=0\)

Exercise \(\PageIndex{36}\)

\((1-2x^3)y''+6x^2y'+24xy=0\)

Answer

$$p(r)=-2r(r-1)+6r+24=-2(r-6)(r+2)$$;
(A) $$\displaystyle\prod_{j=0}^{m-1}{p(3j)\over3j+2}=\prod_{j=0}^{m-1}
(-6)(j-2)$$.

(B) $$\displaystyle\prod_{j=0}^{m-1}{p(3j+1)\over3j+4}=
\prod_{j=0}^{m-1}{(-6)(j+1)(3j-5)\over3j+4}=
(-1)^m6^mm\prod_{j=0}^{m-1}{3j-5\over3j+4}$$.
Substituting (A) and (B) into the
result of Exercise~7.2.32\part{c} yields

$$y=\displaystyle{a_0(1-4x^3+4x^6)+a_1\sum_{m=0}^\infty
2^m\left[\prod_{j=0}^{m-1}{3j-5\over3j+4}\right]x^{3m+1}}$$.

Exercise \(\PageIndex{37}\)

\((1-x^3)y''+15x^2y'-63xy=0\)

Exercise \(\PageIndex{38}\)

Consider the equation

\begin{equation}\label{eq:3.2E.10}
\left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0,
\end{equation}

where \(k\) is a positive integer, and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\).

(a) Modify the argument used to prove Theorem \((3.2.2)\) to show that

\begin{eqnarray*}
y=\sum_{n=0}^\infty a_nx^n
\end{eqnarray*}

is a solution of \eqref{eq:3.2E.10} if and only if \(a_n=0\) for \(2\le n\le k+1\) and

\begin{eqnarray*}
a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.
\end{eqnarray*}

(b) Show from (a) that \(a_n=0\) unless \(n=(k+2)m\) or \(n=(k+2)m+1\) for some nonnegative integer \(m\), and that

\begin{eqnarray*}
a_{(k+2)(m+1)}&=&-{p\left((k+2)m\right)\over (k+2)(m+1)[(k+2)(m+1)-1]}a_{(k+2)m},\quad m\ge 0, \\
\mbox{and}\\
a_{(k+2)(m+1)+1}&=&-{p\left((k+2)m+1\right)\over[(k+2)(m+1)+1](k+2)(m+1)} a_{(k+2)m+1},\quad m\ge0,
\end{eqnarray*}

where \(a_0\) and \(a_1\) may be specified arbitrarily.

(c) Conclude from (b) that the power series in \(x\) for the general solution of \eqref{eq:3.2E.10} is

\begin{eqnarray*}
\begin{array}{l}
y=a_0\displaystyle{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p\left((k+2)j\right)\over(k+2)(j+1)-1}\right] {x^{(k+2)m}\over(k+2)^m m!}}\\
\quad+a_1\displaystyle{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p\left((k+2)j+1\right)\over(k+2)(j+1)+1}\right] {x^{(k+2)m+1}\over(k+2)^mm!}}.
\end{array}
\end{eqnarray*}

Answer

a. Let $$Ly=(1+\alpha x^{k+2})y''+\beta x^{k+1}y'+\gamma x^ky$$. If
$$y=\displaystyle\sum_{n=0}^\infty a_nx^n$$, then
$$Ly=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=0}^\infty
p(n)a_nx^{n+k}=
\sum_{n=-k}^{-1}(n+k+2)(n+k-1)a_{n+k+2}x^{n+k}+\sum_{n=0}^\infty
[(n+k+2)(n+k+1)a_{n+k+2}+p(n)a_n]x^{n+k}
= 0$$ if and only if $$a_k=0$$ for $$2\le n\le k+1$$ and (A) $$\displaystyle
a_{n+k+1}=-{p(n)\over(n+k+2)(n+k+1)}a_n$$ for $$n\ge0$$.

b.  If \(a_n=0\) then \(a_{n+(k+2)m}=0\) for all \(m \ge 0\), from
(A).
 

In Exercises \((3.2E.39)\) to \((3.2E.44)\), use the method of Exercise \((3.2E.38)\) to find the power series in \(x\) for the general solution.

Exercise \(\PageIndex{39}\)

\((1+2x^5)y''+14x^4y'+10x^3y=0\)

Exercise \(\PageIndex{40}\)

\(y''+x^2y=0\)

Answer

\(k=2\) and
\(p(r)=1\);
(A) $$\displaystyle\prod_{j=0}^{m-1}{p(4j)\over4j+3}=
{1\over\prod_{j=0}^{m-1}(4j+3)}$$;
(B) $$\displaystyle\prod_{j=0}^{m-1}{p(4j+1)\over(4j+5)}=
{1\over\prod_{j=0}^{m-1}4j+5}$$.
Substituting (A) and (B) into the
result of Exercise~3.2.38\part{c} yields

$$y=\displaystyle{a_0\sum_{m=0}^\infty
(-1)^m{x^{4m}\over4^mm!\prod_{j=0}^{m-1}(4j+3)}+a_1\sum_{m=0}^\infty
(-1)^m{x^{4m+1}\over4^mm!\prod_{j=0}^{m-1}(4j+5)}}$$.
 

Exercise \(\PageIndex{41}\)

\(y''+x^6y'+7x^5y=0\)

Exercise \(\PageIndex{42}\)

\((1+x^8)y''-16x^7y'+72x^6y=0\)

Answer

\(k=6\) and
$$p(r)=r(r-1)-16r+72=(r-9)(r-8) $$;
(A) $$\displaystyle\prod_{j=0}^{m-1}{p(8j)\over8j+7}=
\prod_{j=0}^{m-1}{8(j-1)(8j-9)\over8j+7}$$;

(B) $$\displaystyle\prod_{j=0}^{m-1}{p(8j+1)\over(8j+9)}=
\prod_{j=0}^{m-1}{8(j-1)(8j-7)\over8j+9}$$;

Substituting (A) and (B) into the
result of Exercise~3.2.38\part{c} yields

$$y=\displaystyle{a_0\left(1-{9\over7}x^8\right)+a_1\left(x-{7\over9}x^9\right)}$$.
 

Exercise \(\PageIndex{43}\)

\((1-x^6)y''-12x^5y'-30x^4y=0\)

Exercise \(\PageIndex{44}\)

\(y''+x^5y'+6x^4y=0\)

Answer

  \(k=4\) and
\(p(r)=r+6\);
(A) $$\displaystyle\prod_{j=0}^{m-1}{p(6j)\over6j+5}=
\prod_{j=0}^{m-1}{6(j+1)\over6j+5}={6^mm!\over\prod_{j=0}^{m-1}
(6j+5)}$$;

(B) $$\displaystyle\prod_{j=0}^{m-1}{p(6j+1)\over(6j+7)}=1$$;

Substituting (A) and (B) into the
result of Exercise~7.2.38\part{c} yields

$$y=\displaystyle{a_0\sum_{m=0}^\infty(-1)^m{x^{6m}\over\prod_{j=0}^{m-1}(6j+5)}+
a_1\sum_{m=0}^\infty(-1)^m{x^{6m+1}\over6^mm!}}$$.