8.3.1: Series Solutions About an Ordinary Point (Exercises)
- Page ID
- 103527
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In Exercises 1-11 find the series in \(x\) for the general solution.
1. \((1+x^2)y''+6xy'+6y=0\)
2. \((1+x^2)y''+2xy'-2y=0\)
3. \((1+x^2)y''-8xy'+20y=0\)
4. \((1-x^2)y''-8xy'-12y=0\)
5. \((1+x^2)y''-10xy'+28y=0\)
6. \(y''-xy=0\)
7. \(y''-2xy'+y=0\)
8. \(y''+x^2y'+xy=0\)
9. \((x-1)y''+y'=0\)
10. \(y''-(x+1)y'-y=0\)
11. \((x^2+2)y''+3xy'-y=0\)
In Exercises 12-23 find the power series solution \(y=\sum _{n=0}^{\infty} a_{n}x^{n}\) of the initial value problem.
12. \((1+x^2)y''+xy'+y=0,\quad y(0)=2,\quad y'(0)=-1\)
13. \((1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1\)
14. \((1+8x^2)y''+2y=0,\quad y(0)=2,\quad y'(0)=-1\)
15. \((1+3x)y''+xy'+2y=0,\quad y(0)=2,\quad y'(0)=-3\)
16. \((1+x+2x^2)y''+(2+8x)y'+4y=0,\quad y(0)=-1,\quad y'(0)=2\)
17. \((1-2x^2)y''+(2-6x)y'-2y=0,\quad y(0)=1,\quad y'(0)=0\)
18. \((1+x+3x^2)y''+(2+15x)y'+12y=0,\quad y(0)=0,\quad y'(0)=1\)
19. \((2+x)y''+(1+x)y'+3y=0,\quad y(0)=4,\quad y'(0)=3\)
20. \((3+3x+x^2)y''+(6+4x)y'+2y=0,\quad y(0)=7,\quad y'(0)=3\)
21. \((4+x)y''+(2+x)y'+2y=0,\quad y(0)=2,\quad y'(0)=5\)
22. \((x-1)y''-xy'+y=0,\quad y(0)=-2,\quad y'(0)=6\)
23. \(y''-2xy'+8y=0,\quad y(0)=3,\quad y'(0)=0\)
In Exercises 24 and 25 find the power series in \(x-x_{0}\) for the general solution.
24. \(y''-y=0;\quad x_0=3\)
25. \((11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2\)
In Exercises 26-63 find the power series solution \(y=\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) for the solution of the initial value problem. Take \(x_{0}\) to be the point where the initial conditions are imposed.
26. \((x^2-4)y''-xy'-3y=0,\quad y(0)=-1,\quad y'(0)=2\)
27. \(y''+(x-3)y'+3y=0,\quad y(3)=-2,\quad y'(3)=3\)
28. \((5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1\)
29. \((4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6\)
30. \((x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4\)
31. \((2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3\)
32. \((2-3x+2x^2)y''-(4-6x)y'+2y=0,\quad y(1)=1,\quad y'(1)=-1\)
33. \((3x+2x^2)y''+10(1+x)y'+8y=0,\quad y(-1)=1,\quad y'(-1)=-1\)
34. \((1-x+x^2)y''-(1-4x)y'+2y=0,\quad y(1)=2,\quad y'(1)=-1\)
35. \((2+x)y''+(2+x)y'+y=0,\quad y(-1)=-2,\quad y'(-1)=3\)
36. \(x^2y''-(6-7x)y'+8y=0,\quad y(1)=1,\quad y'(1)=-2\)
37. \((2+4x)y''-4y'-(6+4x)y=0,\quad y(0)=2,\quad y'(0)=-7\)
38. \((1+2x)y''-(1-2x)y'-(3-2x)y=0,\quad y(1)=1,\quad y'(1)=-2\)
39. \((5+2x)y''-y'+(5+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)
40. \((4+x)y''-(4+2x)y'+(6+x)y=0,\quad y(-3)=2,\quad y'(-3)=-2\)
41. \((2+3x)y''-xy'+2xy=0,\quad y(0)=-1,\quad y'(0)=2\)
42. \((3+2x)y''+3y'-xy=0,\quad y(-1)=2,\quad y'(-1)=-3\)
43. \((3+2x)y''-3y'-(2+x)y=0,\quad y(-2)=-2,\quad y'(-2)=3\)
44. \((10-2x)y''+(1+x)y=0,\quad y(2)=2,\quad y'(2)=-4\)
45. \((7+x)y''+(8+2x)y'+(5+x)y=0,\quad y(-4)=1,\quad y'(-4)=2\)
46. \((6+4x)y''+(1+2x)y=0,\quad y(-1)=-1,\quad y'(-1)=2\)
47. \(y''+2xy'+(3+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)
48. \(y''-3xy'+(5+2x^2)y=0,\quad y(0)=1,\quad y'(0)=-2\)
49. \(y''+5xy'-(3-x^2)y=0,\quad y(0)=6,\quad y'(0)=-2\)
50. \(y''-2xy'-(2+3x^2)y=0,\quad y(0)=2,\quad y'(0)=-5\)
51. \(y''-3xy'+(2+4x^2)y=0,\quad y(0)=3,\quad y'(0)=6\)
52. \(2y''+5xy'+(4+2x^2)y=0,\quad y(0)=3,\quad y'(0)=-2\)
53. \(3y''+2xy'+(4-x^2)y=0,\quad y(0)=-2,\quad y'(0)=3\)
54. \((1+x)y''+x^2y'+(1+2x)y=0,\quad y(0)-2,\quad y'(0)=3\)
55. \(y''+(1+2x+x^2)y'+2y=0,\quad y(0)=2,\quad y'(0)=3\)
56. \((1+x^2)y''+(2+x^2)y'+xy=0,\quad y(0)=-3,\quad y'(0)=5\)
57. \((1+x)y''+(1-3x+2x^2)y'-(x-4)y=0,\quad y(1)=-2,\quad y'(1)=3\)
58. \(y''+(13+12x+3x^2)y'+(5+2x)y=0,\quad y(-2)=2,\quad y'(-2)=-3\)
59. \((1+2x+3x^2)y''+(2-x^2)y'+(1+x)y=0,\quad y(0)=1,\quad y'(0)=-2\)
60. \((3+4x+x^2)y''-(5+4x-x^2)y'-(2+x)y=0,\quad y(-2)=2,\quad y'(-2)=-1\)
61. \((1+2x+x^2)y''+(1-x)y=0,\quad y(0)=2,\quad y'(0)=-1\)
62. \((x-2x^2)y''+(1+3x-x^2)y'+(2+x)y=0,\quad y(1)=1,\quad y'(1)=0\)
63. \((16-11x+2x^2)y''+(10-6x+x^2)y'-(2-x)y=0,\quad y(3)=1,\quad y'(3)=-2\)
64. a. Find a power series in \(x\) for the general solution of \[(1+x^2)y''+4xy'+2y=0. \tag{A}\]
b. 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 \((-1,1)\).
c. Show that the expression obtained in (b) is actually the general solution of of (A) on \((-\infty,\infty)\).
65. 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.
In Exercises 66 and 67 find the power series in \(x\) for the general solution.
66. \(y''-xy=1\)
67. \(y''+xy'-y=e^x\); where \(e^x=\sum_{n=0}^{\infty} {x^n\over n!}\)