
# 7.2E: Series Solutions Near an Ordinary Point I (Exercises)

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In Exercises [exer:7.2.1} –[exer:7.2.8} find the power series in $$x$$ for the general solution.

[exer:7.2.1] $$(1+x^2)y''+6xy'+6y=0$$

[exer:7.2.2] $$(1+x^2)y''+2xy'-2y=0$$

[exer:7.2.3] $$(1+x^2)y''-8xy'+20y=0$$

[exer:7.2.4] $$(1-x^2)y''-8xy'-12y=0$$

[exer:7.2.5] $$(1+2x^2)y''+7xy'+2y=0$$

[exer:7.2.6] $${(1+x^2)y''+2xy'+{1\over4}y=0}$$

[exer:7.2.7] $$(1-x^2)y''-5xy'-4y=0$$

[exer:7.2.8] $$(1+x^2)y''-10xy'+28y=0$$

[exer:7.2.9] Find the power series in $$x$$ for the general solution of $$y''+xy'+2y=0$$.

For several choices of $$a_0$$ and $$a_1$$, use differential equations software to solve the initial value problem $y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$ numerically on $$(-5,5)$$.

For fixed $$r$$ in $$\{1,2,3,4,5\}$$ graph $T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$

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

[exer:7.2.10] Follow the directions of Exercise [exer:7.2.9} for the differential equation $y''+2xy'+3y=0.\nonumber$

[exer:7.2.11] $$(1+x^2)y''+xy'+y=0,\quad y(0)=2,\quad y'(0)=-1$$

[exer:7.2.12] $$(1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1$$

[exer:7.2.13] $$(1+8x^2)y''+2y=0,\quad y(0)=2,\quad y'(0)=-1$$

[exer:7.2.14] Do the next experiment for various choices of real numbers $$a_0$$, $$a_1$$, and $$r$$, with $$0<r<1/\sqrt2$$.

Use differential equations software to solve the initial value problem

$(1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$

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

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

$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$ and the solution obtained in (a) on $$(-r,r)$$. Continue increasing $$N$$ until there’s no perceptible difference between the two graphs.

[exer:7.2.15] Do (a) and (b) for several values of $$r$$ in $$(0,1)$$:

Use differential equations software to solve the initial value problem $(1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1, \eqno{\rm(A)}\nonumber$ numerically on $$(-r,r)$$.

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

$T_N(x)=\sum_{n=0}^N a_nx^n\nonumber$

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

Try (a) and (b) with $$r=1.2$$. Explain your results.

[exer:7.2.16] $$y''-y=0;\quad x_0=3$$

[exer:7.2.17] $$y''-(x-3)y'-y=0;\quad x_0=3$$

[exer:7.2.18] $$(1-4x+2x^2)y''+10(x-1)y'+6y=0;\quad x_0=1$$

[exer:7.2.19] $$(11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2$$

[exer:7.2.20] $$(5+6x+3x^2)y''+9(x+1)y'+3y=0;\quad x_0=-1$$

[exer:7.2.21] $$(x^2-4)y''-xy'-3y=0,\quad y(0)=-1,\quad y'(0)=2$$

[exer:7.2.22] $$y''+(x-3)y'+3y=0,\quad y(3)=-2,\quad y'(3)=3$$

[exer:7.2.23] $$(5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1$$

[exer:7.2.24] $$(4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6$$

[exer:7.2.25] $$(x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4$$

[exer:7.2.26] $$(2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3$$

[exer:7.2.27] Find a power series in $$x$$ for the general solution of $(1+x^2)y''+4xy'+2y=0. \eqno{\rm (A)}\nonumber$

Use (a) and the formula ${1\over1-r}=1+r+r^2+\cdots+r^n+\cdots \quad(-1<r<1)\nonumber$ for the sum of a geometric series to find a closed form expression for the general solution of (A) on $$(-1,1)$$.

Show that the expression obtained in (b) is actually the general solution of of (A) on $$(-\infty,\infty)$$.

[exer:7.2.28] Use Theorem [thmtype:7.2.2} to show that the power series in $$x$$ for the general solution of $(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber$

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

[exer:7.2.29] Use Exercise [exer:7.2.28} to show that all solutions of $(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber$

are polynomials if and only if $\alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1),\nonumber$

where $$r$$ and $$s$$ are nonnegative integers.

[exer:7.2.30] Use Exercise [exer:7.2.28} to show that the power series in $$x$$ for the general solution of $(1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0\nonumber$ is $$y=a_0y_1+a_1y_2$$, where \begin{aligned} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha)(2j+\alpha+2b-1) \right]{x^{2m}\over(2m)!}\\ \noalign{\hspace*{50pt}\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{aligned}\nonumber

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 $(1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0. \eqno{\rm (A)}\nonumber$

Show that (A) implies that $[(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n,\nonumber$ and use this to show that if $$m$$ and $$n$$ are nonnegative integers, then $\begin{array}{ll} [(1-x^2)^bP_n']'P_m-[(1-x^2)^bP_m']'P_n=&\\[4pt] \hspace*{40pt}\left[m(m+2b-1)-n(n+2b-1)\right](1-x^2)^{b-1}P_mP_n.& \end{array} \eqno{\rm (B)}\nonumber$

Now suppose $$b>0$$. Use (B) and integration by parts to show that if $$m\ne n$$, then $\int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0.\nonumber$ (We say that $$P_m$$ and $$P_n$$ are orthogonal on $$(-1,1)$$ with respect to the weighting function$$(1-x^2)^{b-1}$$.)

[exer:7.2.31] Use Exercise [exer:7.2.28} to show that the power series in $$x$$ for the general solution of Hermite’s equation $y''-2xy'+2\alpha y=0\nonumber$ is $$y=a_0y_1+a_1y_1$$, where \begin{aligned} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha) \right]{2^mx^{2m}\over(2m)!}\\ \noalign{\hspace*{25pt}\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{aligned}\nonumber

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 $P_n''-2xP_n'+2nP_n=0. \eqno{\rm (A)}\nonumber$

Show that (A) implies that $[e^{-x^2}P_n']'=-2ne^{-x^2}P_n,\nonumber$ and use this to show that if $$m$$ and $$n$$ are nonnegative integers, then $[e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n. \eqno{\rm (B)}\nonumber$

Use (B) and integration by parts to show that if $$m\ne n$$, then $\int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0.\nonumber$

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

[exer:7.2.32] Consider the equation $\left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0, \eqno{\rm (A)}\nonumber$ and let $$p(n)=\alpha n(n-1)+\beta n+\gamma$$. (The special case $$y''-xy=0$$ of (A) is Airy’s equation.)

Modify the argument used to prove Theorem [thmtype:7.2.2} to show that $y=\sum_{n=0}^\infty a_nx^n\nonumber$ is a solution of (A) if and only if $$a_2=0$$ and $a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.\nonumber$

Show from (a) that $$a_n=0$$ unless $$n=3m$$ or $$n=3m+1$$ for some nonnegative integer $$m$$, and that \begin{aligned} a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0,\\ \noalign{\hspace*{50pt}\mbox{and}}\\ a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0,\end{aligned}\nonumber

where $$a_0$$ and $$a_1$$ may be specified arbitrarily.

Conclude from (b) that the power series in $$x$$ for the general solution of (A) is $\begin{array}{l} y={a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\[4pt] \qquad{+a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p(3j+1)\over3j+4}\right] {x^{3m+1}\over3^mm!}}. \end{array}\nonumber$

[exer:7.2.33] $$y''-xy=0$$

[exer:7.2.34] $$(1-2x^3)y''-10x^2y'-8xy=0$$

[exer:7.2.35] $$(1+x^3)y''+7x^2y'+9xy=0$$ [exer:7.2.36] $$(1-2x^3)y''+6x^2y'+24xy=0$$

[exer:7.2.37] $$(1-x^3)y''+15x^2y'-63xy=0$$

[exer:7.2.38] Consider the equation $\left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0, \eqno{\rm (A)}\nonumber$ where $$k$$ is a positive integer, and let $$p(n)=\alpha n(n-1)+\beta n+\gamma$$.

Modify the argument used to prove Theorem [thmtype:7.2.2} to show that $y=\sum_{n=0}^\infty a_nx^n\nonumber$ is a solution of (A) if and only if $$a_n=0$$ for $$2\le n\le k+1$$ and $a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.\nonumber$

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{aligned} 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, \\ \noalign{\hspace*{50pt}\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{aligned}\nonumber

where $$a_0$$ and $$a_1$$ may be specified arbitrarily.

Conclude from (b) that the power series in $$x$$ for the general solution of (A) is

$\begin{array}{l} y=a_0{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p\left((k+2)j\right)\over(k+2)(j+1)-1}\right] {x^{(k+2)m}\over(k+2)^m m!}}\\ [15pt] \qquad+a_1{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p\left((k+2)j+1\right)\over(k+2)(j+1)+1}\right] {x^{(k+2)m+1}\over(k+2)^mm!}}. \end{array}\nonumber$

[exer:7.2.39] $$(1+2x^5)y''+14x^4y'+10x^3y=0$$

[exer:7.2.40] $$y''+x^2y=0$$ [exer:7.2.41] $$y''+x^6y'+7x^5y=0$$

[exer:7.2.42] $$(1+x^8)y''-16x^7y'+72x^6y=0$$

[exer:7.2.43] $$(1-x^6)y''-12x^5y'-30x^4y=0$$

[exer:7.2.44] $$y''+x^5y'+6x^4y=0$$