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7.3.1: Series Solutions Near an Ordinary Point II (Exercises)

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    30752
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    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 \(0<r<1/\sqrt2\).

    1. 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.)
    2. 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 \(0<r<2\).

    1. 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?)
    2. 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 \(r>0\).

    1. 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.)
    2. 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 \(a_1\).

    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 \((-r,r)\).
    2. 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.

    1. 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 \[P_{0}(x)=1+\alpha x+\beta x^{2}\nonumber \]
    2. 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 \((-\rho,\rho)\).
    3. 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 \((-\rho,\rho)\).
    4. 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 \(1/r_2\).
    5. 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 \((-\rho,\rho)\).
    6. 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 \((-\rho,\rho)\).
    7. 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. \((1+3x+2x^2)y''+(6+8x)y'+4y=0\)
    2. \((1-5x+6x^2)y''-(10-24x)y'+12y=0\)
    3. \((1-4x+4x^2)y''-(8-16x)y'+8y=0\)
    4. \((4+4x+x^2)y''+(8+4x)y'+2y=0\)
    5. \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

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


    This page titled 7.3.1: Series Solutions Near an Ordinary Point II (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.