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

Last updated

Jul 20, 2020
Save as PDF
- 7.3: Series Solutions Near an Ordinary Point II
- Prelude to Series Solutions of Linear Second Order Equations

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\id}{\mathrm{id}}$ $\newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$ $\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$ $\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\id}{\mathrm{id}}$

$\newcommand{\Span}{\mathrm{span}}$

$\newcommand{\kernel}{\mathrm{null}\,}$

$\newcommand{\range}{\mathrm{range}\,}$

$\newcommand{\RealPart}{\mathrm{Re}}$

$\newcommand{\ImaginaryPart}{\mathrm{Im}}$

$\newcommand{\Argument}{\mathrm{Arg}}$

$\newcommand{\norm}[1]{\| #1 \|}$

$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$\newcommand{\Span}{\mathrm{span}}$ $\newcommand{\AA}{\unicode[.8,0]{x212B}}$

$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vectorC}[1]{\textbf{#1}}$

$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$

$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$

$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$

$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$

$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$

$\newcommand{\avec}{\mathbf a}$

$\newcommand{\bvec}{\mathbf b}$

$\newcommand{\cvec}{\mathbf c}$

$\newcommand{\dvec}{\mathbf d}$

$\newcommand{\dtil}{\widetilde{\mathbf d}}$

$\newcommand{\evec}{\mathbf e}$

$\newcommand{\fvec}{\mathbf f}$

$\newcommand{\nvec}{\mathbf n}$

$\newcommand{\pvec}{\mathbf p}$

$\newcommand{\qvec}{\mathbf q}$

$\newcommand{\svec}{\mathbf s}$

$\newcommand{\tvec}{\mathbf t}$

$\newcommand{\uvec}{\mathbf u}$

$\newcommand{\vvec}{\mathbf v}$

$\newcommand{\wvec}{\mathbf w}$

$\newcommand{\xvec}{\mathbf x}$

$\newcommand{\yvec}{\mathbf y}$

$\newcommand{\zvec}{\mathbf z}$

$\newcommand{\rvec}{\mathbf r}$

$\newcommand{\mvec}{\mathbf m}$

$\newcommand{\zerovec}{\mathbf 0}$

$\newcommand{\onevec}{\mathbf 1}$

$\newcommand{\real}{\mathbb R}$

$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$

$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$

$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$

$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$

$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$

$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$

$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$

$\newcommand{\bcal}{\cal B}$

$\newcommand{\ccal}{\cal C}$

$\newcommand{\scal}{\cal S}$

$\newcommand{\wcal}{\cal W}$

$\newcommand{\ecal}{\cal E}$

$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$

$\newcommand{\gray}[1]{\color{gray}{#1}}$

$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$

$\newcommand{\rank}{\operatorname{rank}}$

$\newcommand{\row}{\text{Row}}$

$\newcommand{\col}{\text{Col}}$

$\renewcommand{\row}{\text{Row}}$

$\newcommand{\nul}{\text{Nul}}$

$\newcommand{\var}{\text{Var}}$

$\newcommand{\corr}{\text{corr}}$

$\newcommand{\len}[1]{\left|#1\right|}$

$\newcommand{\bbar}{\overline{\bvec}}$

$\newcommand{\bhat}{\widehat{\bvec}}$

$\newcommand{\bperp}{\bvec^\perp}$

$\newcommand{\xhat}{\widehat{\xvec}}$

$\newcommand{\vhat}{\widehat{\vvec}}$

$\newcommand{\uhat}{\widehat{\uvec}}$

$\newcommand{\what}{\widehat{\wvec}}$

$\newcommand{\Sighat}{\widehat{\Sigma}}$

$\newcommand{\lt}{<}$

$\newcommand{\gt}{>}$

$\newcommand{\amp}{&}$

$\definecolor{fillinmathshade}{gray}{0.9}$

Q7.3.1

In Exercises 7.3.1-7.3.12 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.

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

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

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

4. $(1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1$

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

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

7. $(4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5$

8. $(2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1$

9. $(3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1$

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

11. $(2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3$

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

Q7.3.2

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 $(1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A}$ numerically on $(-r,r)$ . (See Example 7.3.1.)
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.

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 $(3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1, \tag{A}$ 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.

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 $y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \tag{A}$ 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.

16. Do the following experiment for several choices of $a_0$ and $a_1$ .

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

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

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

Q7.3.3

In Exercises 7.3.19-7.3.28 find the coefficients $a_{0},...a_{N}$ for $N$ at least $7$ in the series solution $y=\sum_{n=0}^{\infty}a_{n}(x-x_{0})^{n}\nonumber$ of the initial value problem. Take $x_{0}$ to be the point where the initial conditions are imposed.

19. $(2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7$

20. $(1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2$

21. $(5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$

22. $(4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2$

23. $(2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2$

24. $(3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3$

25. $(3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3$

26. $(10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4$

27. $(7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2$

28. $(6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2$

Q7.3.4

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 $a_{n+2}=-{\gamma+\alpha n\over n+2}\,a_{n+1}-{\beta n(n-1)+\delta n+\epsilon\over(n+2)(n+1)}\, a_n.\nonumber$

30.

Let $\alpha$ and $\beta$ be constants, with $\beta\ne0$ . Show that $y=\sum_{n=0}^\infty a_nx^n$ is a solution of $(1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0 \tag{A}$ if and only if $a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0. \tag{B}$ 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 $P_{0}(x)=1+\alpha x+\beta x^{2}\nonumber$
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 $1/r_1$ .

31. Use the results of Exercise 7.3.30 to find the general solution of the given equation on any interval on which polynomial multiplying $y''$ has no zeros.

$(1+3x+2x^2)y''+(6+8x)y'+4y=0$
$(1-5x+6x^2)y''-(10-24x)y'+12y=0$
$(1-4x+4x^2)y''-(8-16x)y'+8y=0$
$(4+4x+x^2)y''+(8+4x)y'+2y=0$
$(4+8x+3x^2)y''+(16+12x)y'+6y=0$

Q7.3.5

In Exercises 7.3.32-7.3.38 find the coefficients $a_{0}, ..., a_{N}$ for $N$ at least $7$ in the series solution $y=\sum_{n=0}^{\infty} a_{n}x^{n}$ of the initial value problem.

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

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

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

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

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

37. $2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2$

38. $3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3$

Q7.3.6

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.

Q7.3.7

In Exercises 7.3.40-7.3.49 find the coefficients $a_{0}, ..., a_{N}$ for $N$ at least $7$ in the series solution $y=\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\nonumber$ of the initial value problem. Take $x_{0}$ to be the point where the initial conditions are imposed.

40. $(1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3$

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

42. $(1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5$

43. $(1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3$

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

45. $(1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2$

46. $(3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1$

47. $(1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1$

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

49. $(16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y,\quad y(3)=1,\quad y'(3)=-2$

Search

Text Color

Text Size

Margin Size

Font Type

Q7.3.1

Q7.3.2

Q7.3.3

Q7.3.4

Q7.3.5

Q7.3.6

Q7.3.7

Q7.3.1

Q7.3.2

Q7.3.3

Q7.3.4

Q7.3.5

Q7.3.6

Q7.3.7

Support Center

How can we help?