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Mathematics LibreTexts

3.7E: Exercises

  • Page ID
    17856
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    In Exercises \((3.7E.1)\) to \((3.7E.40)\), find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

    Exercise \(\PageIndex{1}\)

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

    Answer

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    Exercise \(\PageIndex{2}\)

    \(xy''+y=0\)

    Answer

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    Exercise \(\PageIndex{3}\)

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

    Answer

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    Exercise \(\PageIndex{4}\)

    \(xy''+xy'+y=0\)

    Answer

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    Exercise \(\PageIndex{5}\)

    \(2x^2(2+3x)y''+x(4+21x)y'-(1-9x)y=0\)

    Answer

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    Exercise \(\PageIndex{6}\)

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

    Answer

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    Exercise \(\PageIndex{7}\)

    \(4x^2y''+4xy'-(9-x)y=0\)

    Answer

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    Exercise \(\PageIndex{8}\)

    \(x^2y''+10xy'+(14+x)y=0\)

    Answer

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    Exercise \(\PageIndex{9}\)

    \(4x^2(1+x)y''+4x(3+8x)y'-(5-49x)y=0\)

    Answer

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    Exercise \(\PageIndex{10}\)

    \(x^2(1+x)y''-x(3+10x)y'+30xy=0\)

    Answer

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    Exercise \(\PageIndex{11}\)

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

    Answer

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    Exercise \(\PageIndex{12}\)

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

    Answer

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    Exercise \(\PageIndex{13}\)

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

    Answer

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    Exercise \(\PageIndex{14}\)

    \(x^2(1+2x)y''+x(9+13x)y'+(7+5x)y=0\)

    Answer

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    Exercise \(\PageIndex{15}\)

    \(4x^2y''-2x(4-x)y'-(7+5x)y=0\)

    Answer

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    Exercise \(\PageIndex{16}\)

    \(3x^2(3+x)y''-x(15+x)y'-20y=0\)

    Answer

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    Exercise \(\PageIndex{17}\)

    \(x^2(1+x)y''+x(1-10x)y'-(9-10x)y=0\)

    Answer

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    Exercise \(\PageIndex{18}\)

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

    Answer

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    Exercise \(\PageIndex{19}\)

    \(x^2(1+2x)y''-2x(3+14x)y'+(6+100x)y=0\)

    Answer

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    Exercise \(\PageIndex{20}\)

    \(x^2(1+x)y''-x(6+11x)y'+(6+32x)y=0\)

    Answer

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    Exercise \(\PageIndex{21}\)

    \(4x^2(1+x)y''+4x(1+4x)y'-(49+27x)y=0\)

    Answer

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    Exercise \(\PageIndex{22}\)

    \(x^2(1+2x)y''-x(9+8x)y'-12xy=0\)

    Answer

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    Exercise \(\PageIndex{23}\)

    \(x^2(1+x^2)y''-x(7-2x^2)y'+12y=0\)

    Answer

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    Exercise \(\PageIndex{24}\)

    \(x^2y''-x(7-x^2)y'+12y=0\)

    Answer

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    Exercise \(\PageIndex{25}\)

    \(xy''-5y'+xy=0\)

    Answer

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    Exercise \(\PageIndex{26}\)

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

    Answer

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    Exercise \(\PageIndex{27}\)

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

    Answer

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    Exercise \(\PageIndex{28}\)

    \(4x^2y''+2x(8+x^2)y'+(5+3x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{29}\)

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

    Answer

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    Exercise \(\PageIndex{30}\)

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

    Answer

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    Exercise \(\PageIndex{31}\)

    \(4x^2y''+8xy'-(35-x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{32}\)

    \(9x^2y''-3x(11+2x^2)y'+(13+10x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{33}\)

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

    Answer

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    Exercise \(\PageIndex{34}\)

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

    Answer

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    Exercise \(\PageIndex{35}\)

    \(x^2(1+x^2)y''+x(5+11x^2)y'+24x^2y=0\)

    Answer

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    Exercise \(\PageIndex{36}\)

    \(4x^2(1+x^2)y''+8xy'-(35-x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{37}\)

    \(x^2(1+x^2)y''-x(5-x^2)y'-(7+25x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{38}\)

    \(x^2(1+x^2)y''+x(5+2x^2)y'-21y=0\)

    Answer

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    Exercise \(\PageIndex{39}\)

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

    Answer

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    Exercise \(\PageIndex{40}\)

    \(4x^2(1+x^2)y''+4x(2+x^2)y'-(15+x^2)y=0\)

    Answer

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    Exercise \(\PageIndex{41}\)

    (a) Under the assumptions of Theorem \((3.7.1)\), show that

    \begin{eqnarray*}
    y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n
    \end{eqnarray*}

    and

    \begin{eqnarray*}
    y_2=x^{r_2}\sum_{n=0}^{k-1}a_n(r_2)x^n+C\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)
    \end{eqnarray*}

    are linearly independent.

    Hint: Show that if \(c_1\) and \(c_2\) are constants such that \(c_1y_1+c_2y_2\equiv0\) on an interval \((0,\rho)\), then

    \begin{eqnarray*}
    x^{-r_2}(c_1y_1(x)+c_2y_2(x))=0,\quad 0<x<\rho.
    \end{eqnarray*}

    Then let \(x\to0+\) to conclude that \(c_2\)=0.

    (b) Use the result of part (a) to complete the proof of Theorem \((3.7.1)\).

    Answer

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    Exercise \(\PageIndex{42}\)

    Find a fundamental set of Frobenius solutions of Bessel's equation

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

    in the case where \(\nu\) is a positive integer.

    Answer

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    Exercise \(\PageIndex{43}\)

    Prove Theorem \((3.7.2)\).

    Answer

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    Exercise \(\PageIndex{44}\)

    Under the assumptions of Theorem \((3.7.1)\), show that \(C=0\) if and only if \(p_1(r_2+ \ell)=0\) for some integer \(\ell\) in \(\{0,1,\dots,k-1\}\).

    Answer

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    Exercise \(\PageIndex{45}\)

    Under the assumptions of Theorem \((3.7.2)\), show that \(C=0\) if and only if \(p_2(r_2+2\ell)=0\) for some integer \(\ell\) in \(\{0,1,\dots,k-1\}\).

    Answer

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    Exercise \(\PageIndex{46}\)

    Let

    \begin{eqnarray*}
    Ly=\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y
    \end{eqnarray*}

    and define

    \begin{eqnarray*}
    p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0.
    \end{eqnarray*}

    Show that if

    \begin{eqnarray*}
    p_0(r)=\alpha_0(r-r_1)(r-r_2)
    \end{eqnarray*}

    where \(r_1-r_2=k\), a positive integer, then \(Ly=0\) has the solutions

    \begin{eqnarray*}
    y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+k)}\left(\gamma_1\over\alpha_0\right)^n x^n
    \end{eqnarray*}

    and

    \begin{eqnarray*}
    y_2&=&x^{r_2}\sum_{n=0}^{k-1} {(-1)^n\over n!\prod_{j=1}^n(j-k)} \left(\gamma_1\over\alpha_0\right)^n x^n\\
    &&-{1\over k!(k-1)!}\left(\gamma_1\over\alpha_0\right)^k\left(y_1\ln x- x^{r_1}\sum_{n=1}^\infty {(-1)^n\over n!\prod_{j=1}^n(j+k)}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{2j+k\over j(j+k)}\right)x^n\right).
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{47}\)

    Let

    \begin{eqnarray*}
    Ly=\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_2x^2)y
    \end{eqnarray*}

    and define

    \begin{eqnarray*}
    p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0.
    \end{eqnarray*}

    Show that if

    \begin{eqnarray*}
    p_0(r)=\alpha_0(r-r_1)(r-r_2)
    \end{eqnarray*}

    where \(r_1-r_2=2k\), an even positive integer, then \(Ly=0\) has the solutions

    \begin{eqnarray*}
    y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over 4^mm!\prod_{j=1}^m(j+k)}\left(\gamma_2\over\alpha_0\right)^m x^{2m}
    \end{eqnarray*}

    and

    \begin{eqnarray*}
    y_2&=&x^{r_2}\sum_{m=0}^{k-1} {(-1)^m\over4^mm!\prod_{j=1}^m(j-k)} \left(\gamma_2\over\alpha_0\right)^m x^{2m}\\
    &&-{2\over 4^kk!(k-1)!}\left(\gamma_2\over\alpha_0\right)^k\left(y_1\ln x- {x^{r_1}\over2}\sum_{m=1}^\infty {(-1)^m\over 4^mm!\prod_{j=1}^m(j+k)}\left(\gamma_2\over\alpha_0\right)^m \left(\sum_{j=1}^m{2j+k\over j(j+k)}\right)x^{2m}\right).
    \end{eqnarray*}

    Answer

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    Exercise \(\PageIndex{48}\)

    Let \(L\) be as in Exercises \((3.5E.57)\) and \((3.5E.58)\), and suppose the indicial polynomial of \(Ly=0\) is

    \begin{eqnarray*}
    p_0(r)=\alpha_0(r-r_1)(r-r_2),
    \end{eqnarray*}

    with \(k=r_1-r_2\), where \(k\) is a positive integer. Define \(a_0(r)=1\) for all \(r\). If \(r\) is a real number such that \(p_0(n+r)\)
    is nonzero for all positive integers \(n\), define

    \begin{eqnarray*}
    a_n(r)=-{1\over p_0(n+r)}\sum_{j=1}^n p_j(n+r-j)a_{n-j}(r),\,n\ge1,
    \end{eqnarray*}

    and let

    \begin{eqnarray*}
    y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n.
    \end{eqnarray*}

    Define

    \begin{eqnarray*}
    a_n(r_2)=-{1\over p_0(n+r_2)}\sum_{j=1}^n p_j(n+r_2-j)a_{n-j}(r_2)\mbox{ if } n\ge1\mbox{ and }n\ne k,
    \end{eqnarray*}

    and let \(a_k(r_2)\) be arbitrary.

    (a) Conclude from Exercise \((3.6E.66)\) that

    \begin{eqnarray*}
    L\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)=k\alpha_0x^{r_1}.
    \end{eqnarray*}

    (b) Conclude from Exercise \((3.5E.57)\) that

    \begin{eqnarray*}
    L\left(x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n\right)=Ax^{r_1},
    \end{eqnarray*}

    where

    \begin{eqnarray*}
    A=\sum_{j=1}^k p_j(r_1-j)a_{k-j}(r_2).
    \end{eqnarray*}

    (c) Show that \(y_1\) and

    \begin{eqnarray*}
    y_2=x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n -{A\over k\alpha_0} \left(y_1 \ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)
    \end{eqnarray*}

    form a fundamental set of Frobenius solutions of \(Ly=0\).

    (d) Show that choosing the arbitrary quantity \(a_k(r_2)\) to be nonzero merely adds a multiple of \(y_1\) to \(y_2\). Conclude that we may as well take \(a_k(r_2)=0\).

    Answer

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