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Mathematics LibreTexts

3.2E: Exercises

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    17553
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    In Exercises \((3.2E.1)\) to \((3.2E.8)\), find the power series in \(x\) for the general solution.

    Exercise \(\PageIndex{1}\)

    \((1+x^2)y''+6xy'+6y=0\)

    Answer

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    Exercise \(\PageIndex{2}\)

    \((1+x^2)y''+2xy'-2y=0\)

    Answer

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    Exercise \(\PageIndex{3}\)

    \((1+x^2)y''-8xy'+20y=0\)

    Answer

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    Exercise \(\PageIndex{4}\)

    \((1-x^2)y''-8xy'-12y=0\)

    Answer

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    Exercise \(\PageIndex{5}\)

    \((1+2x^2)y''+7xy'+2y=0\)

    Answer

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    Exercise \(\PageIndex{6}\)

    \(\displaystyle{(1+x^2)y''+2xy'+{1\over4}y=0}\)

    Answer

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    Exercise \(\PageIndex{7}\)

    \((1-x^2)y''-5xy'-4y=0\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \((1+x^2)y''-10xy'+28y=0\)

    Answer

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    Exercise \(\PageIndex{9}\)

    (a) Find the power series in \(x\) for the general solution of \(y''+xy'+2y=0\).

    (b) For several choices of \(a_0\) and \(a_1\), use differential equations software to solve the initial value problem

    \begin{equation}\label{eq:3.2E.1}
    y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
    \end{equation}

    numerically on \((-5,5)\).

    (c) For fixed \(r\) in \(\{1,2,3,4,5\}\) graph

    \begin{eqnarray*}
    T_N(x)=\sum_{n=0}^N a_nx^n
    \end{eqnarray*}

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

    Answer

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    Exercise \(\PageIndex{10}\)

    Follow the directions of Exercise \((3.2E.9)\) for the differential equation

    \begin{eqnarray*}
    y''+2xy'+3y=0.
    \end{eqnarray*}

    Answer

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    In Exercises \((3.2E.11)\) to \((3.2E.13)\), find \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

    Exercise \(\PageIndex{11}\)

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

    Answer

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    Exercise \(\PageIndex{12}\)

    \((1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1\)

    Answer

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    Exercise \(\PageIndex{13}\)

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

    Answer

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    Exercise \(\PageIndex{14}\)

    Do the next experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<1/\sqrt2\).

    (a) Use differential equations software to solve the initial value problem

    \begin{equation}\label{eq:3.2E.2}
    (1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
    \end{equation}

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

    (b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.2E.2}, and graph

    \begin{eqnarray*}
    T_N(x)=\sum_{n=0}^N a_nx^n
    \end{eqnarray*}

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

    Answer

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    Exercise \(\PageIndex{15}\)

    Do part (a) and part (b) for several values of \(r\) in \((0,1)\):

    (a) Use differential equations software to solve the initial value problem

    \begin{equation}\label{eq:3.2E.3}
    (1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1,
    \end{equation}

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

    (b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.2E.3}, and graph

    \begin{eqnarray*}
    T_N(x)=\sum_{n=0}^N a_nx^n
    \end{eqnarray*}

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

    (c) Try part (a) and part (b) with \(r=1.2\). Explain your results.

    Answer

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    In Exercises \((3.2E.16)\) to \((3.2E.20)\), find the power series in \(-x_0\) for the general solution.

    Exercise \(\PageIndex{16}\)

    \(y''-y=0;\quad x_0=3\)

    Answer

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    Exercise \(\PageIndex{17}\)

    \(y''-(x-3)y'-y=0;\quad x_0=3\)

    Answer

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    Exercise \(\PageIndex{18}\)

    \((1-4x+2x^2)y''+10(x-1)y'+6y=0;\quad x_0=1\)

    Answer

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    Exercise \(\PageIndex{19}\)

    \((11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2\)

    Answer

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    Exercise \(\PageIndex{20}\)

    \((5+6x+3x^2)y''+9(x+1)y'+3y=0;\quad x_0=-1\)

    Answer

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    In Exercises \((3.2E.21)\) to \((3.2E.26)\), find \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the power series \(y=\sum_{n=0}^\infty a_n(x-x_0)^n\) for the solution of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

    Exercise \(\PageIndex{21}\)

    \((x^2-4)y''-xy'-3y=0,\quad y(0)=-1,\quad y'(0)=2\)

    Answer

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    Exercise \(\PageIndex{22}\)

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

    Answer

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    Exercise \(\PageIndex{23}\)

    \((5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1\)

    Answer

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    Exercise \(\PageIndex{24}\)

    \((4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6\)

    Answer

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    Exercise \(\PageIndex{25}\)

    \((x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4\)

    Answer

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    Exercise \(\PageIndex{26}\)

    \((2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3\)

    Answer

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    Exercise \(\PageIndex{27}\)

    (a) Find a power series in \(x\) for the general solution of

    \begin{equation}\label{eq:3.2E.4}
    (1+x^2)y''+4xy'+2y=0.
    \end{equation}

    (b) Use (a) and the formula

    \begin{eqnarray*}
    {1\over1-r}=1+r+r^2+\cdots+r^n+\cdots \quad(-1<r<1)
    \end{eqnarray*}

    for the sum of a geometric series to find a closed form expression for the general solution of \eqref{eq:3.2E.4} on \((-1,1)\).

    (c) Show that the expression obtained in part (b) is actually the general solution of \eqref{eq:3.2E.4} on \((-\infty,\infty)\).

    Answer

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    Exercise \(\PageIndex{28}\)

    Use Theorem \((3.2.2)\) to show that the power series in \(x\) for the general solution of

    \begin{eqnarray*}
    (1+\alpha x^2)y''+\beta xy'+\gamma y=0
    \end{eqnarray*}

    is

    \begin{eqnarray*}
    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)!}.
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{29}\)

    Use Exercise \((3.2E.28)\) to show that all solutions of

    \begin{eqnarray*}
    (1+\alpha x^2)y''+\beta xy'+\gamma y=0
    \end{eqnarray*}

    are polynomials if and only if

    \begin{eqnarray*}
    \alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1),
    \end{eqnarray*}

    where \(r\) and \(s\) are nonnegative integers.

    Answer

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    Exercise \(\PageIndex{30}\)

    (a) Use Exercise \((3.2E.28)\) to show that the power series in \(x\) for the general solution of

    \begin{eqnarray*}
    (1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0
    \end{eqnarray*}

    is \(y=a_0y_1+a_1y_2\), where

    \begin{eqnarray*}
    y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha)(2j+\alpha+2b-1) \right]{x^{2m}\over(2m)!}\\
    \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{eqnarray*}

    (b) 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

    \begin{equation}\label{eq:3.2E.5}
    (1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0.
    \end{equation}

    (c) Show that \eqref{eq:3.2E.5} implies that

    \begin{eqnarray*}
    [(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n,
    \end{eqnarray*}

    and use this to show that if \(m\) and \(n\) are nonnegative integers, then

    \begin{equation}\label{eq:3.2E.6}
    \begin{array}{ll}
    [(1-x^2)^bP_n']'P_m-[(1-x^2)^bP_m']'P_n=&\\
    \left[m(m+2b-1)-n(n+2b-1)\right](1-x^2)^{b-1}P_mP_n.&
    \end{array}
    \end{equation}

    (d) Now suppose \(b>0\). Use \eqref{eq:3.2E.6} and integration by parts to show that if \(m\ne n\), then

    \begin{eqnarray*}
    \int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0.
    \end{eqnarray*}

    (We say that \(P_m\) and \(P_n\) are \( \textcolor{blue}{\mbox{orthogonal on}} \) \((-1,1)\) \( \textcolor{blue}{\mbox{with respect to the weighting function}} \) \((1-x^2)^{b-1}\).)
    \end{alist}

    Answer

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    Exercise \(\PageIndex{31}\)

    (a) Use Exercise \((3.2E.28)\) to show that the power series in \(x\) for the general solution of Hermite's equation

    \begin{eqnarray*}
    y''-2xy'+2\alpha y=0
    \end{eqnarray*}

    is \(y=a_0y_1+a_1y_1\), where

    \begin{eqnarray*}
    y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha) \right]{2^mx^{2m}\over(2m)!}\\
    \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{eqnarray*}

    (b) 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

    \begin{equation}\label{eq:3.2E.7}
    P_n''-2xP_n'+2nP_n=0.
    \end{equation}

    (c) Show that \eqref{eq:3.2E.7} implies that

    \begin{eqnarray*}
    [e^{-x^2}P_n']'=-2ne^{-x^2}P_n,
    \end{eqnarray*}

    and use this to show that if \(m\) and \(n\) are nonnegative integers, then

    \begin{equation}\label{eq:3.2E.8}
    [e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n.
    \end{equation}

    (d) Use \eqref{eq:3.2E.8} and integration by parts to show that if \(m\ne n\), then

    \begin{eqnarray*}
    \int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0.
    \end{eqnarray*}

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

    Answer

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    Exercise \(\PageIndex{32}\)

    Consider the equation

    \begin{equation}\label{eq:3.2E.9}
    \left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0,
    \end{equation}

    and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\). (The special case \(y''-xy=0\) of \eqref{eq:3.2E.9} is Airy's equation.)

    (a) Modify the argument used to prove Theorem \((3.2.2)\) to show that

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty a_nx^n
    \end{eqnarray*}

    is a solution of \eqref{eq:3.2E.9} if and only if \(a_2=0\) and

    \begin{eqnarray*}
    a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.
    \end{eqnarray*}

    (b) Show from (a) that \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\), and that

    \begin{eqnarray*}
    a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0,\\
    \mbox{and}\\
    a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0,
    \end{eqnarray*}

    where \(a_0\) and \(a_1\) may be specified arbitrarily.

    (c) Conclude from (b) that the power series in \(x\) for the general solution of \eqref{eq:3.2E.9} is

    \begin{eqnarray*}
    \begin{array}{l}
    y=\displaystyle{a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\
    \qquad\displaystyle{+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}
    \end{eqnarray*}

    Answer

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    In Exercises \((3.2E.33)\) to \((3.2E.37)\), use the method of Exercise \((3.2E.32)\) to find the power series in \(x\) for the general solution.

    Exercise \(\PageIndex{33}\)

    \(y''-xy=0\)

    Answer

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    Exercise \(\PageIndex{34}\)

    \((1-2x^3)y''-10x^2y'-8xy=0\)

    Answer

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    Exercise \(\PageIndex{35}\)

    \((1+x^3)y''+7x^2y'+9xy=0\)

    Answer

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    Exercise \(\PageIndex{36}\)

    \((1-2x^3)y''+6x^2y'+24xy=0\)

    Answer

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    Exercise \(\PageIndex{37}\)

    \((1-x^3)y''+15x^2y'-63xy=0\)

    Answer

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    Exercise \(\PageIndex{38}\)

    Consider the equation

    \begin{equation}\label{eq:3.2E.10}
    \left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0,
    \end{equation}

    where \(k\) is a positive integer, and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\).

    (a) Modify the argument used to prove Theorem \((3.2.2)\) to show that

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty a_nx^n
    \end{eqnarray*}

    is a solution of \eqref{eq:3.2E.10} if and only if \(a_n=0\) for \(2\le n\le k+1\) and

    \begin{eqnarray*}
    a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.
    \end{eqnarray*}

    (b) 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{eqnarray*}
    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, \\
    \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{eqnarray*}

    where \(a_0\) and \(a_1\) may be specified arbitrarily.

    (c) Conclude from (b) that the power series in \(x\) for the general solution of \eqref{eq:3.2E.10} is

    \begin{eqnarray*}
    \begin{array}{l}
    y=a_0\displaystyle{\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!}}\\
    \quad+a_1\displaystyle{\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}
    \end{eqnarray*}

    Answer

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    In Exercises \((3.2E.39)\) to \((3.2E.44)\), use the method of Exercise \((3.2E.38)\) to find the power series in \(x\) for the general solution.

    Exercise \(\PageIndex{39}\)

    \((1+2x^5)y''+14x^4y'+10x^3y=0\)

    Answer

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    Exercise \(\PageIndex{40}\)

    \(y''+x^2y=0\)

    Answer

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    Exercise \(\PageIndex{41}\)

    \(y''+x^6y'+7x^5y=0\)

    Answer

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    Exercise \(\PageIndex{42}\)

    \((1+x^8)y''-16x^7y'+72x^6y=0\)

    Answer

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    Exercise \(\PageIndex{43}\)

    \((1-x^6)y''-12x^5y'-30x^4y=0\)

    Answer

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    Exercise \(\PageIndex{44}\)

    \(y''+x^5y'+6x^4y=0\)

    Answer

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