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Mathematics LibreTexts

7.2E: Series Solutions Near an Ordinary Point I (Exercises)

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    18310
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    In Exercises [exer:7.2.1} –[exer:7.2.8} find the power series in \(x\) for the general solution.

    [exer:7.2.1] \((1+x^2)y''+6xy'+6y=0\)

    [exer:7.2.2] \((1+x^2)y''+2xy'-2y=0\)

    [exer:7.2.3] \((1+x^2)y''-8xy'+20y=0\)

    [exer:7.2.4] \((1-x^2)y''-8xy'-12y=0\)

    [exer:7.2.5] \((1+2x^2)y''+7xy'+2y=0\)

    [exer:7.2.6] \({(1+x^2)y''+2xy'+{1\over4}y=0}\)

    [exer:7.2.7] \((1-x^2)y''-5xy'-4y=0\)

    [exer:7.2.8] \((1+x^2)y''-10xy'+28y=0\)

    [exer:7.2.9] Find the power series in \(x\) for the general solution of \(y''+xy'+2y=0\).

    For several choices of \(a_0\) and \(a_1\), use differential equations software to solve the initial value problem \[y''+xy'+2y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber \] numerically on \((-5,5)\).

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

    [exer:7.2.10] Follow the directions of Exercise [exer:7.2.9} for the differential equation \[y''+2xy'+3y=0.\nonumber \]

    [exer:7.2.11] \((1+x^2)y''+xy'+y=0,\quad y(0)=2,\quad y'(0)=-1\)

    [exer:7.2.12] \((1+2x^2)y''-9xy'-6y=0,\quad y(0)=1,\quad y'(0)=-1\)

    [exer:7.2.13] \((1+8x^2)y''+2y=0,\quad y(0)=2,\quad y'(0)=-1\)

    [exer:7.2.14] Do the next 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

    \[(1-2x^2)y''-xy'+3y=0,\quad y(0)=a_0,\quad y'(0)=a_1, \eqno{\rm(A)}\nonumber \]

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

    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.

    [exer:7.2.15] Do (a) and (b) for several values of \(r\) in \((0,1)\):

    Use differential equations software to solve the initial value problem \[(1+x^2)y''+10xy'+14y=0,\quad y(0)=5,\quad y'(0)=1, \eqno{\rm(A)}\nonumber \] numerically on \((-r,r)\).

    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. What happens to the required \(N\) as \(r\to1\)?

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

    [exer:7.2.16] \(y''-y=0;\quad x_0=3\)

    [exer:7.2.17] \(y''-(x-3)y'-y=0;\quad x_0=3\)

    [exer:7.2.18] \((1-4x+2x^2)y''+10(x-1)y'+6y=0;\quad x_0=1\)

    [exer:7.2.19] \((11-8x+2x^2)y''-16(x-2)y'+36y=0;\quad x_0=2\)

    [exer:7.2.20] \((5+6x+3x^2)y''+9(x+1)y'+3y=0;\quad x_0=-1\)

    [exer:7.2.21] \((x^2-4)y''-xy'-3y=0,\quad y(0)=-1,\quad y'(0)=2\)

    [exer:7.2.22] \(y''+(x-3)y'+3y=0,\quad y(3)=-2,\quad y'(3)=3\)

    [exer:7.2.23] \((5-6x+3x^2)y''+(x-1)y'+12y=0,\quad y(1)=-1,\quad y'(1)=1\)

    [exer:7.2.24] \((4x^2-24x+37)y''+y=0,\quad y(3)=4,\quad y'(3)=-6\)

    [exer:7.2.25] \((x^2-8x+14)y''-8(x-4)y'+20y=0,\quad y(4)=3,\quad y'(4)=-4\)

    [exer:7.2.26] \((2x^2+4x+5)y''-20(x+1)y'+60y=0,\quad y(-1)=3,\quad y'(-1)=-3\)

    [exer:7.2.27] Find a power series in \(x\) for the general solution of \[(1+x^2)y''+4xy'+2y=0. \eqno{\rm (A)}\nonumber \]

    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 \((-1,1)\).

    Show that the expression obtained in (b) is actually the general solution of of (A) on \((-\infty,\infty)\).

    [exer:7.2.28] Use Theorem [thmtype:7.2.2} to show that the power series in \(x\) for the general solution of \[(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber \]

    is \[y=a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} p(2j)\right] {x^{2m}\over(2m)!} + a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}p(2j+1)\right] {x^{2m+1}\over(2m+1)!}.\nonumber \]

    [exer:7.2.29] Use Exercise [exer:7.2.28} to show that all solutions of \[(1+\alpha x^2)y''+\beta xy'+\gamma y=0\nonumber \]

    are polynomials if and only if \[\alpha n(n-1)+\beta n+\gamma=\alpha(n-2r)(n-2s-1),\nonumber \]

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

    [exer:7.2.30] Use Exercise [exer:7.2.28} to show that the power series in \(x\) for the general solution of \[(1-x^2)y''-2bxy'+\alpha(\alpha+2b-1)y=0\nonumber \] is \(y=a_0y_1+a_1y_2\), where \[\begin{aligned} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha)(2j+\alpha+2b-1) \right]{x^{2m}\over(2m)!}\\ \noalign{\hspace*{50pt}\mbox{and}}\\ y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha)(2j+\alpha+2b) \right]{x^{2m+1}\over(2m+1)!}.\end{aligned}\nonumber \]

    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 \[(1-x^2)P_n''-2bxP_n'+n(n+2b-1)P_n=0. \eqno{\rm (A)}\nonumber \]

    Show that (A) implies that \[[(1-x^2)^b P_n']'=-n(n+2b-1)(1-x^2)^{b-1}P_n,\nonumber \] and use this to show that if \(m\) and \(n\) are nonnegative integers, then \[\begin{array}{ll} [(1-x^2)^bP_n']'P_m-[(1-x^2)^bP_m']'P_n=&\\[5pt] \hspace*{40pt}\left[m(m+2b-1)-n(n+2b-1)\right](1-x^2)^{b-1}P_mP_n.& \end{array} \eqno{\rm (B)}\nonumber \]

    Now suppose \(b>0\). Use (B) and integration by parts to show that if \(m\ne n\), then \[\int_{-1}^1 (1-x^2)^{b-1}P_m(x)P_n(x)\,dx=0.\nonumber \] (We say that \(P_m\) and \(P_n\) are orthogonal on \((-1,1)\) with respect to the weighting function\((1-x^2)^{b-1}\).)

    [exer:7.2.31] Use Exercise [exer:7.2.28} to show that the power series in \(x\) for the general solution of Hermite’s equation \[y''-2xy'+2\alpha y=0\nonumber \] is \(y=a_0y_1+a_1y_1\), where \[\begin{aligned} y_1&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j-\alpha) \right]{2^mx^{2m}\over(2m)!}\\ \noalign{\hspace*{25pt}\mbox{and}}\\ y_2&=&\sum_{m=0}^\infty \left[\prod_{j=0}^{m-1}(2j+1-\alpha) \right]{2^mx^{2m+1}\over(2m+1)!}.\end{aligned}\nonumber \]

    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 \[P_n''-2xP_n'+2nP_n=0. \eqno{\rm (A)}\nonumber \]

    Show that (A) implies that \[[e^{-x^2}P_n']'=-2ne^{-x^2}P_n,\nonumber \] and use this to show that if \(m\) and \(n\) are nonnegative integers, then \[[e^{-x^2}P_n']'P_m-[e^{-x^2}P_m']'P_n= 2(m-n)e^{-x^2}P_mP_n. \eqno{\rm (B)}\nonumber \]

    Use (B) and integration by parts to show that if \(m\ne n\), then \[\int_{-\infty}^\infty e^{-x^2}P_m(x)P_n(x)\,dx=0.\nonumber \]

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

    [exer:7.2.32] Consider the equation \[\left(1+\alpha x^3\right)y''+\beta x^2y'+\gamma xy=0, \eqno{\rm (A)}\nonumber \] and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\). (The special case \(y''-xy=0\) of (A) is Airy’s equation.)

    Modify the argument used to prove Theorem [thmtype:7.2.2} to show that \[y=\sum_{n=0}^\infty a_nx^n\nonumber \] is a solution of (A) if and only if \(a_2=0\) and \[a_{n+3}=-{p(n)\over(n+3)(n+2)}a_n,\quad n\ge0.\nonumber \]

    Show from (a) that \(a_n=0\) unless \(n=3m\) or \(n=3m+1\) for some nonnegative integer \(m\), and that \[\begin{aligned} a_{3m+3}&=&-{p(3m)\over(3m+3)(3m+2)}a_{3m},\quad m\ge 0,\\ \noalign{\hspace*{50pt}\mbox{and}}\\ a_{3m+4}&=&-{p(3m+1)\over(3m+4)(3m+3)} a_{3m+1},\quad m\ge0,\end{aligned}\nonumber \]

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

    Conclude from (b) that the power series in \(x\) for the general solution of (A) is \[\begin{array}{l} y={a_0\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p(3j)\over3j+2}\right] {x^{3m}\over3^m m!}}\\[5pt] \qquad{+a_1\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p(3j+1)\over3j+4}\right] {x^{3m+1}\over3^mm!}}. \end{array}\nonumber \]

    [exer:7.2.33] \(y''-xy=0\)

    [exer:7.2.34] \((1-2x^3)y''-10x^2y'-8xy=0\)

    [exer:7.2.35] \((1+x^3)y''+7x^2y'+9xy=0\) [exer:7.2.36] \((1-2x^3)y''+6x^2y'+24xy=0\)

    [exer:7.2.37] \((1-x^3)y''+15x^2y'-63xy=0\)

    [exer:7.2.38] Consider the equation \[\left(1+\alpha x^{k+2}\right)y''+\beta x^{k+1}y'+\gamma x^ky=0, \eqno{\rm (A)}\nonumber \] where \(k\) is a positive integer, and let \(p(n)=\alpha n(n-1)+\beta n+\gamma\).

    Modify the argument used to prove Theorem [thmtype:7.2.2} to show that \[y=\sum_{n=0}^\infty a_nx^n\nonumber \] is a solution of (A) if and only if \(a_n=0\) for \(2\le n\le k+1\) and \[a_{n+k+2}=-{p(n)\over(n+k+2)(n+k+1)}a_n,\quad n\ge0.\nonumber \]

    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

    \[\begin{aligned} a_{(k+2)(m+1)}&=&-{p\left((k+2)m\right)\over (k+2)(m+1)[(k+2)(m+1)-1]}a_{(k+2)m},\quad m\ge 0, \\ \noalign{\hspace*{50pt}\mbox{and}}\\ a_{(k+2)(m+1)+1}&=&-{p\left((k+2)m+1\right)\over[(k+2)(m+1)+1](k+2)(m+1)} a_{(k+2)m+1},\quad m\ge0,\end{aligned}\nonumber \]

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

    Conclude from (b) that the power series in \(x\) for the general solution of (A) is

    \[\begin{array}{l} y=a_0{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0} {p\left((k+2)j\right)\over(k+2)(j+1)-1}\right] {x^{(k+2)m}\over(k+2)^m m!}}\\ [15pt] \qquad+a_1{\sum^\infty_{m=0}(-1)^m \left[\prod^{m-1}_{j=0}{p\left((k+2)j+1\right)\over(k+2)(j+1)+1}\right] {x^{(k+2)m+1}\over(k+2)^mm!}}. \end{array}\nonumber \]

    [exer:7.2.39] \((1+2x^5)y''+14x^4y'+10x^3y=0\)

    [exer:7.2.40] \(y''+x^2y=0\) [exer:7.2.41] \(y''+x^6y'+7x^5y=0\)

    [exer:7.2.42] \((1+x^8)y''-16x^7y'+72x^6y=0\)

    [exer:7.2.43] \((1-x^6)y''-12x^5y'-30x^4y=0\)

    [exer:7.2.44] \(y''+x^5y'+6x^4y=0\)