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3.3E: Exercises

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    17625
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    In Exercises \((3.3E.1)\) to \((3.3E.12)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

    Exercise \(\PageIndex{1}\)

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

    Exercise \(\PageIndex{2}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((1+x+2x^2)y''+(2+8x)y'+4y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
    +2\sum_{n=2}^\infty n(n-1)a_nx^n
    +2\sum_{n=1}^\infty na_nx^{n-1}
    +8\sum_{n=1}^\infty na_nx^n
    +4\sum_{n=0}^\infty a_nx^n
    =\sum_{n=0}^\infty(n+2)(n+1)(a_{n+2}+a_{n+1}+2a_n)x^n=0\)
    if
    \(a_{n+2}=-a_{n+1}-2a_n, a_n\ge0\).
    Starting with \(a_0=-1\) and \(a_1=2\) yields
    \(y=\displaystyle{-1+2x-4x^3+4x^4+4x^5-12x^6+4x^7+\cdots}\).

    Exercise \(\PageIndex{3}\)

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

    Exercise \(\PageIndex{4}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((1+x+3x^2)y''+(2+15x)y'+12y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
    +3\sum_{n=2}^\infty n(n-1)a_nx^n
    +2\sum_{n=1}^\infty na_nx^{n-1}
    +15\sum_{n=1}^\infty na_nx^n
    +12\sum_{n=0}^\infty a_nx^n
    =\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)(n+2)a_{n+1}+3(n+2)^2a_n]x^n=0\)
    if
    \(a_{n+2}=-a_{n+1}-\displaystyle{3(n+2)\over n+1}a_n\),
    \(a_n\ge0\). Starting with \(a_0=0\) and \(a_1=1\) yields
    \(y=\displaystyle{x-x^2-{7\over2}x^3+{15\over2}x^4+{45\over8}x^5
    -{261\over8}x^6+{207\over16}x^7+\cdots}\).

    Exercise \(\PageIndex{5}\)

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

    Exercise \(\PageIndex{6}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((3+3x+x^2)y''+(6+4x)y'+2y=
    3\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +3\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
    +\sum_{n=2}^\infty n(n-1)a_nx^n
    +6\sum_{n=1}^\infty na_nx^{n-1}
    +4\sum_{n=1}^\infty na_nx^n
    +2\sum_{n=0}^\infty a_nx^n
    =\sum_{n=0}^\infty(n+2)(n+1)[3a_{n+2}+3a_{n+1}+a_n]x^n=0\)
    if
    \(a_{n+2}=-a_{n+1}-a_n/3\),
    \(a_n\ge0\). Starting with \(a_0=7\) and \(a_1=3\) yields
    \(y=\displaystyle{7+3x-{16\over3}x^2+{13\over3}x^3-{23\over9}x^4+{10\over9}x^5
    -{7\over27}x^6-{1\over9}x^7+\cdots}\).

    Exercise \(\PageIndex{7}\)

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

    Exercise \(\PageIndex{8}\)

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

    Answer

     The equation is equivalent to
     \((1+t+2t^2)y''+(2+6t)y'+2y=0\) with \(t=x-1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+t+2t^2)y''+(2+6t)y'+2y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +2\sum_{n=2}^\infty n(n-1)a_nt^n
    +2\sum_{n=1}^\infty na_nt^{n-1}
    +6\sum_{n=1}^\infty na_nt^n
    +2\sum_{n=0}^\infty a_nt^n
    =\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}+(n+2)a_{n+1}+2(n+1)a_n]t^n=0\)
    Thus,
    \(a_{n+2}=-\displaystyle a_{n+1}-{2(n+1)\over n+2}a_n\),
    \(a_n\ge0\). Starting with \(a_0=1\) and \(a_1=-1\) yields
    \(y=\displaystyle{1-(x-1)+{4\over3}(x-1)^3-{4\over3}(x-1)^4-{4\over5}(x-1)^5
    +{136\over45}(x-1)^6-{104\over63}(x-1)^7+\cdots}\)

    Exercise \(\PageIndex{9}\)

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

    Exercise \(\PageIndex{10}\)

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

    Answer

    The equation is equivalent to
    \((1+t+t^2)y''+(3+4t)y'+2y=0\) with \)t=x-1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+t+t^2)y''+(3+4t)y'+2y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +\sum_{n=2}^\infty n(n-1)a_nt^n
    +3\sum_{n=1}^\infty na_nt^{n-1}
    +4\sum_{n=1}^\infty na_nt^n
    +2\sum_{n=0}^\infty a_nt^n
    =\sum_{n=0}^\infty(n+1)[(n+2)a_{n+2}+(n+3)a_{n+1}+(n+2)a_n]t^n=0\)
    if
    \(a_{n+2}=-\displaystyle{n+3\over n+2}a_{n+1}-a_n\),
    \(a_n\ge0\). Starting with \(a_0=2\) and \(a_1=-1\) yields
    \(y=\displaystyle{2-(x-1)-{1\over2}(x-1)^2+{5\over3}(x-1)^3-{19\over12}(x-1)^4
    +{7\over30}(x-1)^5+{59\over45}(x-1)^6-{1091\over630}(x-1)^7+\cdots}\)

    Exercise \(\PageIndex{11}\)

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

    Exercise \(\PageIndex{12}\)

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

    Answer

     The equation is equivalent to
     \((1+2t+t^2)y''+(1+7t)y'+8y=0\) with \(t=x-1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+2t+t^2)y''+(1+7t)y'+8y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +\sum_{n=2}^\infty n(n-1)a_nt^n
    +\sum_{n=1}^\infty na_nt^{n-1}
    +7\sum_{n=1}^\infty na_nt^n
    +8\sum_{n=0}^\infty a_nt^n
    =\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)(2n+1)a_{n+1}+(n+2)(n+4)a_n]t^n=0\)
    if
    \(a_{n+2}=-\displaystyle{2n+1\over n+2}a_{n+1}-{n+4\over n+1}a_n\),
    \(a_n\ge0\). Starting with \(a_0=1\) and \(a_1=-2\) yields
    \(y=\displaystyle{1-2(x-1)-3(x-1)^2+8(x-1)^3-4(x-1)^4-{42\over5}(x-1)^5
    +19(x-1)^6-{604\over35}(x-1)^7+\cdots}\)

    Exercise \(\PageIndex{13}\)

    Do the following 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:3.3E.1}
    (1+x+2x^2)y''+(1+7x)y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
    \end{equation}

    numerically on \((-r,r)\). (See Example \((3.3.1)\).)

    (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:3.3E.1}, and graph

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

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

    Exercise \(\PageIndex{14}\)

    Do the following experiment for various choices of real numbers \(a_0\), \(a_1\), and \(r\), with \(0<r<2\).

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

    \begin{equation}\label{eq:3.3E.2}
    (3+x)y''+(1+2x)y'-(2-x)y=0,\quad y(-1)=a_0,\quad y'(-1)=a_1,
    \end{equation}

    numerically on \((-1-r,-1+r)\). (See Example \((3.3.2)\). Why this interval?)

    (b) For \(N=2\), \(3\), \(4\), \(\dots\), compute \(a_2,\dots,a_N\) in the power series solution

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty a_n(x+1)^n
    \end{eqnarray*}

    of \eqref{eq:3.3E.2}, and graph

    \begin{eqnarray*}
    T_N(x)=\sum_{n=0}^N a_n(x+1)^n
    \end{eqnarray*}

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

    Exercise \(\PageIndex{15}\)

    Do the following experiment for several choices of \(a_0\), \(a_1\), and \(r\), with \(r>0\).

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

    \begin{equation}\label{eq:3.3E.3}
    y''+3xy'+(4+2x^2)y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
    \end{equation}

    numerically on \((-r,r)\). (See Example \((3.3.3)\).)

    (b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^\infty a_nx^n\) of \eqref{eq:3.3E.3}, and graph

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

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

    Exercise \(\PageIndex{16}\)

    Do the following experiment for several choices of \(a_0\) and \(a_1\).

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

    \begin{equation}\label{eq:3.3E.4}
    (1-x)y''-(2-x)y'+y=0,\quad y(0)=a_0,\quad y'(0)=a_1,
    \end{equation}

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

    (b) Find the coefficients \(a_0\), \(a_1\), \(\dots\), \(a_N\) in the power series solution \(y=\sum_{n=0}^Na_nx^n\) of \eqref{eq:3.3E.4}, and graph

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

    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 \(r\to1\)?

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((1-x)y''-(2-x)y'+y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    -\sum_{n=2}^\infty n(n-1)a_nx^{n-1}
    -2\sum_{n=1}^\infty na_nx^{n-1}
    +\sum_{n=1}^\infty na_nx^n
    +\sum_{n=0}^\infty a_nx^n
    =\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-(n+2)(n+1)a_{n+1}+(n+1)a_n]x^n=0\)
    if
    \(a_{n+2}=a_{n+1}-\displaystyle{a_n\over n+2}, a_n\ge0\).

    Exercise \(\PageIndex{17}\)

    Follow the directions of Exercise \((3.3E.16)\) for the initial value problem

    \begin{eqnarray*}
    (1+x)y''+3y'+32y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
    \end{eqnarray*}

    Exercise \(\PageIndex{18}\)

    Follow the directions of Exercise \((3.3E.16)\) for the initial value problem

    \begin{eqnarray*}
    (1+x^2)y''+y'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1.
    \end{eqnarray*}

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((1+x^2)y''+y'+2y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nx^n
    +\sum_{n=1}^\infty na_nx^{n-1}
    +2\sum_{n=0}^\infty a_nx^n
    =\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}+(n+1)a_{n+1}+(n^2-n+2)a_n]x^n=0\)
    if
     \(a_{n+2}=-\displaystyle{1\over n+2}a_{n+1}-
    \displaystyle{n^2-n+2\over(n+2)(n+1)}a_n\).

    In Exercises \((3.3E.19)\) to \((3.3E.28)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty a_n(x-x_0)^n
    \end{eqnarray*}

    of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

    Exercise \(\PageIndex{19}\)

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

    Exercise \(\PageIndex{20}\)

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

    Answer

     The equation is equivalent to
    \((3+2t)y''+(1+2t)y'-(1-2t)y=0\) with \(t=x-1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((3+2t)y''+(1+2t)y'-(1-2t)y=
    3\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +\sum_{n=1}^\infty na_nt^{n-1}
    +2\sum_{n=1}^\infty na_nt^n
    -\sum_{n=0}^\infty a_nt^n
    +2\sum_{n=0}^\infty a_nt^{n+1}
    =(6a_2+a_1-a_0)+
    \sum_{n=1}^\infty[3(n+2)(n+1)a_{n+2}+(n+1)(2n+1)a_{n+1}+
    (2n-1)a_n+2a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{a_1-a_0\over6}\) and
    \(a_{n+2}=-\displaystyle{2n+1\over3(n+2)}a_{n+1}-{2n-1\over3(n+2)(n+1)}a_n
    -{2\over3(n+2)(n+1)}a_{n-1},
    n\ge1\). Starting with \(a_0=1\) and \(a_1=-2\) yields
    \(y=\displaystyle{1-2(x-1)+{1\over2}(x-1)^2-{1\over6}(x-1)^3+{5\over36}(x-1)^4
    -{73\over1080}(x-1)^5+\cdots}\).

    Exercise \(\PageIndex{21}\)

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

    Exercise \(\PageIndex{22}\)

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

    Answer

    The equation is equivalent to
     \((1+t)y''+(2-2t)y'+(3+t)y=0\) with \(t=x+3\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+t)y''+(2-2t)y'+(3+t)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +2\sum_{n=1}^\infty na_nt^{n-1}
    -2\sum_{n=1}^\infty na_nt^n
    +3\sum_{n=0}^\infty a_nt^n
    +\sum_{n=0}^\infty a_nt^{n+1}
    =(2a_2+2a_1+3a_0)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(n+2)(n+1)a_{n+1}
    -(2n-3)a_n+a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{2a_1+3a_0\over2}\) and
    \(a_{n+2}=\displaystyle-a_{n+1}+{(2n-3)a_n-a_{n-1}\over(n+2)(n+1)},
    n\ge1\). Starting with \(a_0=2\) and \(a_1=-2\) yields

    \(y=\displaystyle{2-2(x+3)-(x+3)^2+(x+3)^3-{11\over12}(x+3)^4+
    {67\over60}(x+3)^5+\cdots}\).

    Exercise \(\PageIndex{23}\)

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

    Exercise \(\PageIndex{24}\)

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

    Answer

    The equation is equivalent to
     \((1+2t)y''+3y'+(1-t)y=0\) with \(t=x+1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+2t)y''+3y'+(1-t)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +3\sum_{n=1}^\infty na_nt^{n-1}
    +\sum_{n=0}^\infty a_nt^n
    -\sum_{n=0}^\infty a_nt^{n+1}
    =(2a_2+3a_1+a_0)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(2n+3)(n+1)a_{n+1}
    +a_n-a_{n-1}]t^n=0\)  if \(a_2=-\displaystyle{3a_1+a_0\over2}\) and
    \(a_{n+2}=\displaystyle-{2n+3\over n+2}a_{n+1}-{a_n-a_{n-1}\over(n+2)(n+1)},
    n \ge 1\). Starting with \(a_0=2\) and \(a_1=-3\) yields
    \(y=\displaystyle{2-3(x+1)+{7\over2}(x+1)^2-5(x+1)^3+{197\over24}(x+1)^4
    -{287\over20}(x+1)^5+\cdots}\).

    Exercise \(\PageIndex{25}\)

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

    Exercise \(\PageIndex{26}\)

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

    Answer

     The equation is equivalent to
     \((6-2t)y''+(3+t)y=0\) with \(t=x-2\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((6-2t)y''+(3+t)y=
    6\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    -2\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +3\sum_{n=0}^\infty a_nt^n
    +\sum_{n=0}^\infty a_nt^{n+1}
    =(12a_2+3a_0)+
    \sum_{n=1}^\infty[6(n+2)(n+1)a_{n+2}-2(n+1)na_{n+1}+3a_n
    +a_{n-1}]t^n=0\)  if \)a_2=-\displaystyle{a_0\over4}\) and
    \(a_{n+2}=\displaystyle{n\over3(n+2)}a_{n+1}-{3a_n+a_{n-1}\over6(n+2)(n+1)},
    n\ge1\). Starting with \(a_0=2\) and \(a_1=-4\) yields
    \(y=\displaystyle{2-4(x-2)-{1\over2}(x-2)^2+{2\over9}(x-2)^3+{49\over432}(x-2)^4
    +{23\over1080}(x-2)^5+\cdots}\).

    Exercise \(\PageIndex{27}\)

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

    Exercise \(\PageIndex{28}\)

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

    Answer

     The equation is equivalent to
     \((2+4t)y''-(1-2t)y=0\) with \(t=x+4\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((2+4t)y''-(1-2t)y=
    2\displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +4\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    -\sum_{n=0}^\infty a_nt^n
    +2\sum_{n=0}^\infty a_nt^{n+1}
    =(4a_2-a_0)+
    \sum_{n=1}^\infty[2(n+2)(n+1)a_{n+2}+4(n+1)na_{n+1}
    -a_n+2a_{n-1}]t^n=0\)  if \)a_2=\displaystyle{a_0\over4}\) and
    \(a_{n+2}=\displaystyle-{2n\over
    n+2}a_{n+1}+{a_n-2a_{n-1}\over2(n+2)(n+1)},
    n\ge1\). Starting with \(a_0=-1\) and \(a_1=2\) yields
    \(y=\displaystyle{-1+2(x+1)-{1\over4}(x+1)^2+{1\over2}(x+1)^3-{65\over96}(x+1)^4
    +{67\over80}(x+1)^5+\cdots}\).

    N=5; b=zeros(N,1); b(1)=-1;b(2)=2; b(3)=b(1)/4;
    for n=1:N-2
       b(n+3)=-2*n*b(n+2)/(n+2)+(b(n+1)-2*b(n))/(2*(n+2)*(n+1));
    end

     

    Exercise \(\PageIndex{29}\)

    Show that the coefficients in the power series in \(x\) for the general solution of

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

    satisfy the recurrence relation

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

    Answer

    Let \(Ly=(1+\alpha x+\beta x^2)y''+(\gamma+\delta x)y'+\epsilon y\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \(Ly=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+
    \alpha\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+
    \beta\sum_{n=2}^\infty n(n-1)a_nx^n+
    \gamma\sum_{n=1}^\infty na_nx^{n-1}+
    \delta\sum_{n=1}^\infty na_nx^n+
    \epsilon\sum_{n=0}^\infty a_nx^n=
    \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n+
    \alpha\sum_{n=0}^\infty (n+1)na_{n+1}x^n+
    \beta\sum_{n=0}^\infty n(n-1)a_nx^n+
    \gamma\sum_{n=0}^\infty (n+1)a_{n+1}x^n+
    \delta\sum_{n=0}^\infty na_nx^n+
    \epsilon\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty b_nx^n\),
    where
    \(b_n=(n+1)(n+2)a_{n+2}+(n+1)(\alpha
    n+\gamma)a_{n+1}+[\beta n(n-1)+\delta n+\epsilon]a_n\), which implies
    the conclusion.

    Exercise \(\PageIndex{30}\)

    (a) Let \(\alpha\) and \(\beta\) be constants, with \(\beta\ne0\). Show that \(y=\sum_{n=0}^\infty a_nx^n\) is a solution of

    \begin{equation}\label{eq:3.3E.5}
    (1+\alpha x+\beta x^2)y''+(2\alpha+4\beta x)y'+2\beta y=0
    \end{equation}

    if and only if

    \begin{equation}\label{eq:3.3E.6}
    a_{n+2}+\alpha a_{n+1}+\beta a_n=0,\quad n\ge0.
    \end{equation}

    An equation of this form is called a \( \textcolor{blue}{\mbox{second order homogeneous linear difference equation}} \). The polynomial \(p(r)=r^2+\alpha r+\beta\) is called the \( \textcolor{blue}{\mbox{characteristic polynomial}} \) of \eqref{eq:3.3E.6}. If \(r_1\) and \(r_2\) are the zeros of \(p\), then \(1/r_1\) and \(1/r_2\) are the zeros of

    \begin{eqnarray*}
    P_0(x)=1+\alpha x+\beta x^2.
    \end{eqnarray*}

    (b) 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

    \begin{eqnarray*}
    a_n=c_1r_1^n+c_2r_2^n,\quad n\ge0
    \end{eqnarray*}

    satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty (c_1r_1^n+c_2r_2^n)x^n
    \end{eqnarray*}

    is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

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

    \begin{eqnarray*}
    y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={1\over1-r_2x}
    \end{eqnarray*}

    form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

    (d) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that doesn't contain either \(1/r_1\) or \(1/r_2\).

    (e) 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

    \begin{eqnarray*}
    a_n=(c_1+c_2n)r_1^n,\quad n\ge0
    \end{eqnarray*}

    satisfies \eqref{eq:3.3E.6}. Conclude from this that any function of the form

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty (c_1+c_2n)r_1^nx^n
    \end{eqnarray*}

    is a solution of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

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

    \begin{eqnarray*}
    y_1={1\over1-r_1x}\quad\mbox{ and }\quad y_2={x\over(1-r_1x)^2}
    \end{eqnarray*}

    form a fundamental set of solutions of \eqref{eq:3.3E.5} on \((-\rho,\rho)\).

    (g) Show that \(\{y_1,y_2\}\) is a fundamental set of solutions of \eqref{eq:3.3E.5} on any interval that does not contain \(1/r_1\).

    Answer

    \part{a} Let \(\gamma=2\alpha, \delta=4\beta\), and \(\epsilon=2\beta\)
    in Exercise~3.3.29 to obtain (B).

    \part{b} If \(a_n=c_1r_1^n+c_2r_2^n\), then \(a_{n+2}+\alpha
    a_{n+1}+\beta a_n=c_1r_1^n(r_1^2+\alpha
    r+\beta)+c_2r_2^n(r_2^2+\alpha
    r_2+\beta)=c_1r_1^nP_0(r_1)+c_2r_2^nP_0(r_2)=0\), so \(\{a_n\}\)
    satisfies (B). Since \(1/r_1\) and \(1/r_2\) are the zeros
    of \)P_0\), Theorem~3.2.1 implies  that \(\sum_{n=0}^\infty
    (c_1r_1^n+c_2r_2^n)x^n\) is a solution of (A) on \((-\rho,\rho)\).

    \part{c} If \(|x|<\rho\), then \(|r_1x|<\rho\) and \(|r_2x|<1\), so
    \(\displaystyle\sum_{n=0}^\infty r_i^nx^n={1\over1-r_ix}=y_i, i=1,2\).
    Therefore,
    \part{b} implies that \(\{y_1,y_2\}\) is a fundamental set of solutions
    of (A) on \((-\rho,\rho)\).

    \part{d} (A) can written as \(P_0y''+2P_0'y'+P_0''y=(P_0y)''=0\).
    Therefore, \(P_0y=a+bx\) where \(a\) and \(b\) are arbitrary constants, and
     a partial fraction expansion shows that  the general solution of
    (A)
    on any interval not containing \(1/r_1\) or \(1/r_2\) is \(y=\displaystyle{a+bx\over
    P_0(x)}={c_1\over1-r_1x}+{c_2\over1-r_2x}=c_1y_1+c_2y_2\).

    \part{e} If \(a_n=c_1r_1^n+c_2r_2^n\), then \(a_{n+2}+\alpha
    a_{n+1}+\beta a_n=c_1r_1^n(r_1^2+\alpha
    r+\beta)+c_2r_1^n[(n+2)r_1^2+\alpha(n+1)
    r_2+\beta n)=(c_1+nc_2)r_1^nP_0(r_1)+c_2r_1^nP_0'(r_1)=0\), so
    \)\{a_n\}\) satisfies (B). Since \(1/r_1\) is the only  zero
    of \(P_0\), Theorem~3.2.1 implies  that \(\sum_{n=0}^\infty
    (c_1+c_2n)r_1^n)x^n\) is a solution of (A) on \((-\rho,\rho)\).


    \part{f} If \(|x|<\rho\), then \(|r_1x|<\rho\), so
    \(\displaystyle\sum_{n=0}^\infty r_1^nx^n={1\over1-r_1x}=y_1\).
    Differentiating this and multiplying the result by \(x\)
    shows that
    \(\displaystyle\sum_{n=0}^\infty nr_1^nx^n={r_1x\over(1-r_1x)^2}=r_1y_2\).
    Therefore,
    \part{e} implies that \(\{y_1,y_2\}\) is a fundamental set of solutions
    of (A) on \((-\rho,\rho)\).

    \part{g} The argument is the same as in \part{c}, but now the
    partial fraction expansion can be written as
     \(y=\displaystyle{a+bx\over
    P_0(x)}={c_1\over1-r_1x}+{c_2x\over(1-r_2x)^2}=c_1y_1+c_2y_2\).

    Exercise \(\PageIndex{31}\)

    Use the results of Exercise \((3.3E.30)\) 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) \((4+8x+3x^2)y''+(16+12x)y'+6y=0\)

    In Exercises \((3.3E.32)\) to \((3.3E.38)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution \(y=\sum_{n=0}^\infty a_nx^n\) of the initial value problem.

    Exercise \(\PageIndex{32}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \(y''+2xy'+(3+2x^2)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +2\sum_{n=1}^\infty na_nx^n
    +3\sum_{n=0}^\infty a_nx^n
    +2\sum_{n=0}^\infty a_nx^{n+2}
    =(2a_2+3a_0)+(6a_3+5a_1)x+
    \sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}+(2n+3)a_n+2a_{n-2}]x^n=0\) if
    \(a_2=-3a_0/2, a_3=-5a_1/6\), and
    \(a_{n+2}=-\displaystyle{(2n+3)a_n+2a_{n-2}\over(n+2)(n+1)},
     n\ge2\).
    Starting with \(a_0=1\) and \(a_1=-2\) yields
    \(y=\displaystyle{1-2x-{3\over2}x^2+{5\over3}x^3+{17\over24}x^4-{11\over20}x^5+\cdots}\).

    Exercise \(\PageIndex{33}\)

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

    Exercise \(\PageIndex{34}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \(y''+5xy'-(3-x^2)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +5\sum_{n=1}^\infty na_nx^n
    -3\sum_{n=0}^\infty a_nx^n
    +\sum_{n=0}^\infty a_nx^{n+2}
    =(2a_2-3a_0)+(6a_3+2a_1)x+
    \sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}+(5n-3)a_n+a_{n-2}]x^n=0\) if
    \(a_2=3a_0/2, a_3=-a_1/3\), and
    \(a_{n+2}=-\displaystyle{(5n-3)a_n+a_{n-2}\over(n+2)(n+1)},
     n\ge2\).
    Starting with \(a_0=6\) and \(a_1=-2\) yields
    \(y=\displaystyle{6-2x+9x^2+{2\over3}x^3-{23\over4}x^4-{3\over10}x^5+\cdots}\).

    Exercise \(\PageIndex{35}\)

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

    Exercise \(\PageIndex{36}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \(y''-3xy'+(2+4x^2)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    -3\sum_{n=1}^\infty na_nx^n
    +2\sum_{n=0}^\infty a_nx^n
    +4\sum_{n=0}^\infty a_nx^{n+2}
    =(2a_2+2a_0)+(6a_3-a_1)x+
    \sum_{n=2}^\infty[(n+2)(n+1)a_{n+2}-(3n-2)a_n+4a_{n-2}]x^n=0\) if
    \(a_2=-a_0, a_3=a_1/6\), and
    \(a_{n+2}=\displaystyle{(3n-2)a_n-4a_{n-2}\over(n+2)(n+1)},
     n\ge2\).
    Starting with \(a_0=3\) and \(a_1=6\) yields
    \(y=\displaystyle{3+6x-3x^2+x^3-2x^4-{17\over20}x^5+\cdots}\).

    Exercise \(\PageIndex{37}\)

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

    Exercise \(\PageIndex{38}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \(3y''+2xy'+(4-x^2)y=
    3\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +2\sum_{n=1}^\infty na_nx^n
    +4\sum_{n=0}^\infty a_nx^n
    -\sum_{n=0}^\infty a_nx^{n+2}
    =(6a_2+4a_0)+(18a_3+6a_1)x+
    \sum_{n=2}^\infty[3(n+2)(n+1)a_{n+2}+(2n+4)a_n-a_{n-2}]x^n=0\) if
    \(a_2=-2a_0/3, a_3=-a_1/3\), and
    \(a_{n+2}=-\displaystyle{(2n+4)a_n-a_{n-2}\over3(n+2)(n+1)},
     n\ge2\).
    Starting with \(a_0=-2\) and \(a_1=3\) yields
    \(y=\displaystyle-2+3x+{4\over3}x^2-x^3-{19\over54}x^4+{13\over60}x^5+\cdots\).

    Exercise \(\PageIndex{39}\)

    Find power series in \(x\) for the solutions \(y_1\) and \(y_2\) of

    \begin{eqnarray*}
    y''+4xy'+(2+4x^2)y=0
    \end{eqnarray*}

    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 \((3.3E.40)\) tp \((3.3E.49)\), find the coefficients \(a_0\), \(\dots\), \(a_N\) for \(N\) at least \(7\) in the series solution

    \begin{eqnarray*}
    y=\sum_{n=0}^\infty a_n(x-x_0)^n
    \end{eqnarray*}

    of the initial value problem. Take \(x_0\) to be the point where the initial conditions are imposed.

    Exercise \(\PageIndex{40}\)

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

    Answer

    If \(y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \((1+x)y''+x^2y'+(1+2x)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    + \sum_{n=2}^\infty n(n-1)a_nx^{n-1}
    +\sum_{n=1}^\infty na_nx^{n+1}
    +\sum_{n=0}^\infty a_nx^n
    +2\sum_{n=0}^\infty a_nx^{n+1}
    =(2a_2+a_0)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+(n+1)na_{n+1}+a_n+(n+1)a_{n-1}]x^n=0\)
    if
    \(a_2=-a_0/2\) and
    \(a_{n+2}=-\displaystyle{(n+1)na_{n+1}+a_n+(n+1)a_{n-1}\over(n+2)(n+1)},
     n \ge 1\).
    Starting with \(a_0=-2\) and \(a_1=3\) yields
    \(y=\displaystyle{-2+3x+x^2-{1\over6}x^3-{3\over4}x^4+{31\over120}x^5+\cdots}\).

    Exercise \(\PageIndex{41}\)

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

    Exercise \(\PageIndex{42}\)

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

    Answer

    If \)y=\displaystyle\sum_{n=0}^\infty a_nx^n\), then
    \)(1+x^2)y''+(2+x^2)y'+xy=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nx^{n-2}
    +\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^n
    +2\sum_{n=1}^\infty na_nx^{n-1}
    +\sum_{n=1}^\infty na_nx^{n+1}
    +\sum_{n=1}^\infty a_nx^{n+1}
    =(2a_2+2a_1)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+2(n+1)a_{n+1}+n(n-1)a_n+na_{n-1}]x^n=0\)
    if
    \(a_2=-a_1\) and
    \(a_{n+2}=-\displaystyle{[2(n+1)a_{n+1}+n(n-1)a_n+na_{n-1}]\over(n+2)(n+1)},
     n \ge 1\).
    Starting with \(a_0=-3\) and \(a_1=5\) yields
    \(y=\displaystyle{-3+5x-5x^2+{23\over6}x^3-{23\over12}x^4+{11\over30}x^5+\cdots}\).

    Exercise \(\PageIndex{43}\)

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

    Exercise \(\PageIndex{44}\)

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

    Answer

    The equation is equivalent to
     \(y''+(1+3t^2)y'+(1+2t)y=0\) with \)t=x-2\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \(y''+(1+3t^2)y'+(1+2t)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +\sum_{n=1}^\infty na_nt^{n-1}
    +3\sum_{n=1}^\infty na_nt^{n+1}
    +\sum_{n=0}^\infty a_nt^n
    +2\sum_{n=0}^\infty a_nt^{n+1}=(2a_2+a_1+a_0)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}+
    (n+1)a_{n+1}+a_n+(3n-1)a_{n-1}]t^n=0\) if \)a_2=-(a_1+a_0)/2\) and
    \)a_{n+2}=-\displaystyle{[(n+1)a_{n+1}+a_n+(3n-1)a_{n-1}]\over
    (n+2)(n+1)},
    n \ge 1 \). Starting with \(a_0=2\) and \(a_1=-3\) yields

    \(y=\displaystyle{2-3(x+2)+{1\over2}(x+2)^2-{1\over3}(x+2)^3+{31\over24}(x+2)^4-
    {53\over120}(x+2)^5+\cdots}\).

    Exercise \(\PageIndex{45}\)

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

    Exercise \(\PageIndex{46}\)

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

    Answer

    The equation is equivalent to
    \((1-t^2)y''-(7-8t+t^2)y'+ty=0\) with \)t=x+2\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1-t^2)y''-(7-8t+t^2)y'+ty=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    - \sum_{n=2}^\infty n(n-1)a_nt^n
    -7\sum_{n=1}^\infty na_nt^{n-1}
    +8\sum_{n=1}^\infty na_nt^n
    -\sum_{n=1}^\infty na_nt^{n+1}
    +\sum_{n=0}^\infty a_nt^{n+1}
    =(2a_2-7a_1)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}
    -7(n+1)a_{n+1}-n(n-9)a_n-(n-2)a_{n-1}]t^n=0\) if \)a_2=7a_1/2\)
    and
    \(a_{n+2}=\displaystyle{[7(n+1)a_{n+1}+n(n-9)a_n+(n-2)a_{n-1}]\over
    (n+2)(n+1)},
    n\ge1\). Starting with \(a_0=2\) and \(a_1=-1\) yields

    \(y=\displaystyle{2-(x+2)-{7\over2}(x+2)^2-{43\over6}(x+2)^3-{203\over24}(x+2)^4-{167
    \over30}(x+2)^5+\cdots}\).

    Exercise \(\PageIndex{47}\)

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

    Exercise \(\PageIndex{48}\)

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

    Answer

    The equation is equivalent to
     \((1+3t+2t^2)y''-(3+t-t^2)y'-(3+t)y=0\) with \(t=x-1\).
    If \(y=\displaystyle\sum_{n=0}^\infty a_nt^n\), then
    \((1+3t+2t^2)y''-(3+t-t^2)y'-(3+t)y=
    \displaystyle \sum_{n=2}^\infty n(n-1)a_nt^{n-2}
    +3\sum_{n=2}^\infty n(n-1)a_nt^{n-1}
    +2\sum_{n=2}^\infty n(n-1)a_nt^n
    -3\sum_{n=1}^\infty na_nt^{n-1}
    -\sum_{n=1}^\infty na_nt^n
    +\sum_{n=1}^\infty na_nt^{n+1}
    -3\sum_{n=0}^\infty a_nt^n
    -\sum_{n=0}^\infty a_nt^{n+1}
    =(2a_2-3a_1-3a_0)+
    \sum_{n=1}^\infty[(n+2)(n+1)a_{n+2}
    +3(n^2-1)a_{n+1}+(2n^2-3n-3)(n+1)a_n+(n-2)a_{n-1}]t^n=0\) if
    \(a_2=3(a_1+a_0)/2\) and
    \(a_{n+2}=-\displaystyle{[3(n^2-1)a_{n+1}+(2n^2-3n-3)(n+1)a_n+(n-2)a_{n-1}]\over
    (n+2)(n+1)},
    n\ge1\). Starting with \(a_0=1\) and \(a_1=0\) yields
    \(y=\displaystyle{1+{3\over2}(x-1)^2+{1\over6}(x-1)^3-{1\over8}(x-1)^5+\cdots}\).

    Exercise \(\PageIndex{49}\)

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


    This page titled 3.3E: Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.