
# 7.3E: Series Solutions Near an Ordinary Point II (Exercises)

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In Exercises [exer:7.3.1} –[exer:7.3.12} 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.

[exer:7.3.1] $$(1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3$$

[exer:7.3.2] $$(1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2$$

[exer:7.3.3] $$(1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0$$

[exer:7.3.4] $$(1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1$$

[exer:7.3.5] $$(2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3$$

[exer:7.3.6] $$(3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3$$

[exer:7.3.7] $$(4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5$$

[exer:7.3.8] $$(2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1$$

[exer:7.3.9] $$(3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1$$

[exer:7.3.10] $$(1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1$$

[exer:7.3.11] $$(2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3$$

[exer:7.3.12] $$x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2$$

[exer:7.3.13] Do the following 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

$(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$

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

Example $$\PageIndex{1}$$:

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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:7.3.14] Do the following experiment for various choices of real numbers $$a_0$$, $$a_1$$, and $$r$$, with $$0<r<2$$.

Use differential equations software to solve the initial value problem $(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \eqno{\rm(A)}\nonumber$

numerically on $$(-1-r,-1+r)$$. (See Example 7.3.2}. Why this interval?)

For $$N=2$$, $$3$$, $$4$$, …, compute $$a_2,\dots,a_N$$ in the power series solution

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

of (A), and graph

$T_N(x)=\sum_{n=0}^N a_n(x+1)^n\nonumber$

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

[exer:7.3.15] Do the following experiment for several choices of $$a_0$$, $$a_1$$, and $$r$$, with $$r>0$$.

Use differential equations software to solve the initial value problem $y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber$

numerically on $$(-r,r)$$. (See Example 7.3.3}.)

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

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

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

[exer:7.3.16] Do the following experiment for several choices of $$a_0$$ and $$a_1$$.

Use differential equations software to solve the initial value problem

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

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

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

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

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

[exer:7.3.17] Follow the directions of Exercise [exer:7.3.16} for the initial value problem $(1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber$

[exer:7.3.18] Follow the directions of Exercise [exer:7.3.16} for the initial value problem $(1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.\nonumber$

[exer:7.3.19] $$(2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7$$

[exer:7.3.20] $$(1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2$$

[exer:7.3.21] $$(5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$$

[exer:7.3.22] $$(4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2$$

[exer:7.3.23] $$(2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2$$

[exer:7.3.24] $$(3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3$$

[exer:7.3.25] $$(3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3$$

[exer:7.3.26] $$(10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4$$

[exer:7.3.27] $$(7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2$$

[exer:7.3.28] $$(6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2$$

[exer:7.3.29] Show that the coefficients in the power series in $$x$$ for the general solution of $(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y=0\nonumber$ satisfy the recurrrence relation $a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.\nonumber$

[exer:7.3.30] Let $$\alpha$$ and $$\beta$$ be constants, with $$\beta\ne0$$. Show that $$y=\sum_{n=0}^\infty a_nx^n$$ is a solution of $(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \eqno{\rm (A)}\nonumber$ if and only if $a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \eqno{\rm (B)}\nonumber$

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 $P_0(x)=1+\alpha x+\beta x^2.\nonumber$

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 $a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0\nonumber$ satisfies (B). Conclude from this that any function of the form $y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n\nonumber$ is a solution of (A) on $$(-\rho,\rho)$$.

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

$y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}\nonumber$

form a fundamental set of solutions of (A) on $$(-\rho,\rho)$$.

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 $$1/r_2$$.

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

$a_n=(c_1+c_2n)r_1^n,\quad n\ge0\nonumber$

satisfies (B). Conclude from this that any function of the form

$y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n\nonumber$

is a solution of (A) on $$(-\rho,\rho)$$.

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

$y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}\nonumber$

form a fundamental set of solutions of (A) on $$(-\rho,\rho)$$.

Show that $$\{y_1,y_2\}$$ is a fundamental set of solutions of (A) on any interval that does not contain $$1/r_1$$.

[exer:7.3.31] Use the results of Exercise [exer:7.3.30} 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) $$(4+8x+3x^2)y''+(16+12x)y'+6y=0$$

[exer:7.3.32] $$y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2$$

[exer:7.3.33] $$y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2$$

[exer:7.3.34] $$y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2$$

[exer:7.3.35] $$y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5$$

[exer:7.3.36] $$y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6$$

[exer:7.3.37] $$2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2$$

[exer:7.3.38] $$3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3$$

[exer:7.3.39] Find power series in $$x$$ for the solutions $$y_1$$ and $$y_2$$ of $y''+4xy'+(2+4x^2)y=0\nonumber$ 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.

[exer:7.3.40] $$(1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3$$

[exer:7.3.41] $$y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3$$

[exer:7.3.42] $$(1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5$$

[exer:7.3.43] $$(1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3$$

[exer:7.3.44] $$y''+(13+12x+3x^2)y'+(5+2x),\quad y(-2)=2,\quad y'(-2)=-3$$

[exer:7.3.45] $$(1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2$$

[exer:7.3.46] $$(3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$$

[exer:7.3.47] $$(1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1$$

[exer:7.3.48] $$(x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0$$

[exer:7.3.49] $$(16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2$$