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Mathematics LibreTexts

3.6: The Method of Frobenius II

  • Page ID
    17432
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    In this section we discuss a method for finding two linearly independent Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. As in the preceding section, we consider equations that can be written as

    \begin{equation} \label{eq:3.6.1}
    x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0
    \end{equation}

    where \(\alpha_0\ne0\). We assume that the indicial equation \(p_0(r)=0\) has a repeated real root \(r_1\). In this case Theorem \((3.5.3)\) implies that \eqref{eq:3.6.1} has one solution of the form

    \begin{eqnarray*}
    y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n,
    \end{eqnarray*}

    but does not provide a second solution \(y_2\) such that \(\{y_1,y_2\}\) is a fundamental set of solutions. The following extension of Theorem \((3.5.2)\) provides a way to find a second solution.

    Theorem \(\PageIndex{1}\)

    Let

    \begin{equation} \label{eq:3.6.2}
    Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y,
    \end{equation}

    where \(\alpha_0\ne0\) and define

    \begin{eqnarray*}
    p_0(r)&=&\alpha_0r(r-1)+\beta_0r+\gamma_0,\\
    p_1(r)&=&\alpha_1r(r-1)+\beta_1r+\gamma_1,\\
    p_2(r)&=&\alpha_2r(r-1)+\beta_2r+\gamma_2.\\
    \end{eqnarray*}

    Suppose \(r\) is a real number such that \(p_0(n+r)\) is nonzero for all positive integers \(n\), and define

    \begin{eqnarray*}
    \begin{array}{ccl}
    a_0(r)&=&1,\\
    a_1(r)&=&-\displaystyle{p_1(r)\over p_0(r+1)},\\
    a_n(r)&=&-\displaystyle{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)},\quad n\ge2.
    \end{array}
    \end{eqnarray*}

    Then the Frobenius series

    \begin{equation} \label{eq:3.6.3}
    y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n
    \end{equation}

    satisfies

    \begin{equation} \label{eq:3.6.4}
    Ly(x,r)=p_0(r)x^r.
    \end{equation}

    Moreover,

    \begin{equation} \label{eq:3.6.5}
    {\partial y\over \partial r}(x,r)=y(x,r)\ln x+x^r\sum_{n=1}^\infty a_n'(r) x^n,
    \end{equation}

    and

    \begin{equation} \label{eq:3.6.6}
    L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.
    \end{equation}

    Proof

    Theorem \((3.5.2)\) implies \eqref{eq:3.6.4}. Differentiating formally with respect to \(r\) in \eqref{eq:3.6.3} yields

    \begin{eqnarray*}
    {\partial y\over \partial r}(x,r)&=&\displaystyle{{\partial\over\partial r}(x^r)\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\
    &=&\displaystyle{x^r\ln x\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\
    &=&y(x,r) \ln x + x^r\sum_{n=1}^\infty a_n'(r)x^n,
    \end{eqnarray*}

    which proves \eqref{eq:3.6.5}.

    To prove that \(\partial y(x,r)/\partial r\) satisfies \eqref{eq:3.6.6}, we view \(y\) in \eqref{eq:3.6.2} as a function \(y=y(x,r)\) of two variables, where the prime indicates partial differentiation with respect to \(x\); thus,

    \begin{eqnarray*}
    y'=y'(x,r)={\partial y\over\partial x}(x,r) \quad \mbox{and} \quad y''=y''(x,r)={\partial^2 y\over\partial x^2}(x,r).
    \end{eqnarray*}

    With this notation we can use \eqref{eq:3.6.2} to rewrite \eqref{eq:3.6.4} as

    \begin{equation} \label{eq:3.6.7}
    x^2q_0(x){\partial^2 y\over \partial x^2}(x,r)+xq_1(x){\partial y\over \partial x}(x,r)+q_2(x)y(x,r)=p_0(r)x^r,
    \end{equation}

    where

    \begin{eqnarray*}
    q_0(x)&=&\alpha_0+\alpha_1x+\alpha_2x^2,\\
    q_1(x)&=&\beta_0+\beta_1x+\beta_2x^2,\\
    q_2(x)&=&\gamma_0+\gamma_1x+\gamma_2x^2.\\
    \end{eqnarray*}

    Differentiating both sides of \eqref{eq:3.6.7} with respect to \(r\) yields

    \begin{eqnarray*}
    x^2q_0(x){\partial^3y\over \partial r\partial x^2}(x,r)+ xq_1(x){\partial^2y\over \partial r\partial x}(x,r)+q_2(x){\partial y\over\partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x.
    \end{eqnarray*}

    By changing the order of differentiation in the first two terms on the left we can rewrite this as

    \begin{eqnarray*}
    x^2q_0(x){\partial^3 y\over \partial x^2\partial r}(x,r) +xq_1(x){\partial^2 y\over \partial x\partial r}(x,r)+q_2(x){\partial y\over \partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x,
    \end{eqnarray*}

    or

    \begin{eqnarray*}
    x^2q_0(x){\partial^2\over \partial x^2} \left({\partial y\over\partial r}(x,r)\right) +xq_1(x){\partial\over\partial r}\left({\partial y\over\partial x}(x,r)\right) +q_2(x){\partial y\over\partial r}(x,r)= p'_0(r)x^r+p_0(r) x^r \ln x,
    \end{eqnarray*}

    which is equivalent to \eqref{eq:3.6.6}.

    Theorem \(\PageIndex{2}\)

    Let \(L\) be as in Theorem \((3.6.1)\) and suppose the indicial equation \(p_0(r)=0\) has a repeated real root \(r_1.\) Then

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

    and

    \begin{equation} \label{eq:3.6.8}
    y_2(x)={\partial y\over\partial r}(x,r_1)=y_1(x)\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n
    \end{equation}

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

    Proof

    Since \(r_1\) is a repeated root of \(p_0(r)=0\), the indicial polynomial can be factored as

    \begin{eqnarray*}
    p_0(r)=\alpha_0(r-r_1)^2,
    \end{eqnarray*}

    so

    \begin{eqnarray*}
    p_0(n+r_1)=\alpha_0n^2,
    \end{eqnarray*}

    which is nonzero if \(n>0\). Therefore the assumptions of Theorem \((3.6.1)\) hold with \(r=r_1\), and \eqref{eq:3.6.4} implies that \(Ly_1=p_0(r_1)x^{r_1}=0\). Since

    \begin{eqnarray*}
    p_0'(r)=2\alpha(r-r_1)
    \end{eqnarray*}

    it follows that \(p_0'(r_1)=0\), so \eqref{eq:3.6.6} implies that

    \begin{eqnarray*}
    Ly_2=p_0'(r_1)x^{r_1}+x^{r_1}p_0(r_1)\ln x=0.
    \end{eqnarray*}

    This proves that \(y_1\) and \(y_2\) are both solutions of \(Ly=0\). We leave the proof that \(\{y_1,y_2\}\) is a fundamental set as an exercise (Exercise \((3.6E.53)\)).

    Example \(\PageIndex{1}\)

    Find a fundamental set of solutions of

    \begin{equation} \label{eq:3.6.9}
    x^2(1-2x+x^2)y''-x(3+x)y'+(4+x)y=0.
    \end{equation}

    Compute just the terms involving \(x^{n+r_1}\), where \(0\le n\le4\) and \(r_1\) is the oot of the indicial equation.

    Answer

    For the given equation, the polynomials defined in Theorem \((3.6.1)\) are

    \begin{eqnarray*}
    \begin{array}{lllll}
    p_0(r)&=&r(r-1)-3r+4&=&(r-2)^2,\\
    p_1(r)&=&-2r(r-1)-r+1&=&-(r-1)(2r+1),\\
    p_2(r)&=&r(r-1).
    \end{array}
    \end{eqnarray*}

    Since \(r_1=2\) is a repeated root of the indicial polynomial \(p_0\), Theorem \((3.6.2)\) implies that

    \begin{equation} \label{eq:3.6.10}
    y_1=x^2\sum_{n=0}^\infty a_n(2)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^2\sum_{n=1}^\infty a_n'(2)x^n
    \end{equation}

    form a fundamental set of Frobenius solutions of \eqref{eq:3.6.9}. To find the coefficients in these series, we use the recurrence formulas from Theorem \((3.6.1)\):

    \begin{equation} \label{eq:3.6.11}
    \begin{array}{ccl}
    a_0(r)&=&1,\\
    a_1(r)&=&-\displaystyle{p_1(r)\over p_0(r+1)} =-\displaystyle{(r-1)(2r+1)\over(r-1)^2} =\displaystyle{2r+1\over r-1},\\
    a_n(r)&=&-\displaystyle{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)}\\
    &=&\displaystyle{(n+r-2)\left[(2n+2r-1)a_{n-1}(r) -(n+r-3)a_{n-2}(r)\right]\over(n+r-2)^2}\\
    &=&\displaystyle{{(2n+2r-1)\over(n+r-2)}a_{n-1}(r)- {(n+r-3)\over(n+r-2)}a_{n-2}(r)},n\ge2.
    \end{array}
    \end{equation}

    Differentiating yields

    \begin{equation} \label{eq:3.6.12}
    \begin{array}{ccl}
    a'_1(r)&=&-\displaystyle{3\over (r-1)^2},\\
    a'_n(r)&=&\displaystyle{{2n+2r-1\over n+r-2}a'_{n-1}(r)-{n+r-3\over n+r-2}a'_{n-2}(r)}\\
    &&\displaystyle{-{3\over(n+r-2)^2}a_{n-1}(r)-{1\over(n+r-2)^2}a_{n-2}(r)},\quad n\ge2.
    \end{array}
    \end{equation}

    Setting \(r=2\) in \eqref{eq:3.6.11} and \eqref{eq:3.6.12} yields

    \begin{equation} \label{eq:3.6.13}
    \begin{array}{ccl}
    a_0(2)&=&1,\\
    a_1(2)&=&5,\\
    a_n(2)&=&\displaystyle{{(2n+3)\over n} a_{n-1}(2)-{(n-1)\over n}a_{n-2}(2)},\quad n\ge2
    \end{array}
    \end{equation}

    and

    \begin{equation} \label{eq:3.6.14}
    \begin{array}{ccl}
    a_1'(2)&=&-3,\\
    a'_n(2)&=&\displaystyle{{2n+3\over n}a'_{n-1}(2)-{n-1\over n}a'_{n-2}(2) -{3\over n^2}a_{n-1}(2)-{1\over n^2}a_{n-2}(2)},\quad n\ge2.
    \end{array}
    \end{equation}

    Computing recursively with \eqref{eq:3.6.13} and \eqref{eq:3.6.14} yields

    \begin{eqnarray*}
    a_0(2)=1,\,a_1(2)=5,\,a_2(2)=17,\,a_3(2)={143\over3},\,a_4(2)={355\over3},
    \end{eqnarray*}

    and

    \begin{eqnarray*}
    a_1'(2)=-3,\,a_2'(2)=-{29\over2},\,a_3'(2)=-{859\over18}, \,a_4'(2)=-{4693\over36}.
    \end{eqnarray*}

    Substituting these coefficients into \eqref{eq:3.6.10} yields

    \begin{eqnarray*}
    y_1=x^2\left(1+5x+17x^2+{143\over3}x^3 +{355\over3}x^4+\cdots\right)
    \end{eqnarray*}

    and

    \begin{eqnarray*}
    y_2=y_1 \ln x -x^3\left(3+{29\over2}x+{859\over18}x^2+{4693\over36}x^3 +\cdots\right).
    \end{eqnarray*}

    Since the recurrence formula \eqref{eq:3.6.11} involves three terms, it's not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of \eqref{eq:3.6.9}. However, as we saw in the preceding sections, the recurrence formula for \(\{a_n(r)\}\) involves only two terms if either \(\alpha_1=\beta_1=\gamma_1=0\) or \(\alpha_2=\beta_2=\gamma_2=0\) in \eqref{eq:3.6.1}. In this case, it's often possible to find explicit formulas for the coefficients. The next two examples illustrate this.

    Example \(\PageIndex{2}\)

    Find a fundamental set of Frobenius solutions of

    \begin{equation} \label{eq:3.6.15}
    2x^2(2+x)y''+5x^2y'+(1+x)y=0.
    \end{equation}

    Give explicit formulas for the coefficients in the solutions.

    Answer

    For the given equation, the polynomials defined in Theorem \((3.6.1)\) are

    \begin{eqnarray*}
    \begin{array}{ccccc}
    p_0(r)&=&4r(r-1)+1&=&(2r-1)^2,\\
    p_1(r)&=&2r(r-1)+5r+1&=&(r+1)(2r+1),\\
    p_2(r)&=&0.
    \end{array}
    \end{eqnarray*}

    Since \(r_1=1/2\) is a repeated zero of the indicial polynomial \(p_0\), Theorem \((3.6.2)\) implies that

    \begin{equation} \label{eq:3.6.16}
    y_1=x^{1/2}\sum_{n=0}^\infty a_n(1/2)x^n
    \end{equation}

    and

    \begin{equation} \label{eq:3.6.17}
    y_2=y_1\ln x+x^{1/2}\sum_{n=1}^\infty a_n'(1/2)x^n
    \end{equation}

    form a fundamental set of Frobenius solutions of \eqref{eq:3.6.15}. Since \(p_2\equiv0\), the recurrence formulas in Theorem \((3.6.1)\) reduce to

    \begin{eqnarray*}
    \begin{array}{ccl}
    a_0(r)&=&1,\\
    a_n(r)&=&-\displaystyle{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r),\\
    &=&-\displaystyle{(n+r)(2n+2r-1)\over(2n+2r-1)^2}a_{n-1}(r),\\
    &=&-\displaystyle{n+r\over2n+2r-1}a_{n-1}(r),\quad n\ge0.
    \end{array}
    \end{eqnarray*}

    We leave it to you to show that

    \begin{equation} \label{eq:3.6.18}
    a_n(r)=(-1)^n\prod_{j=1}^n{j+r\over2j+2r-1},\quad n\ge0.
    \end{equation}

    Setting \(r=1/2\) yields

    \begin{equation} \label{eq:3.6.19}
    \begin{array}{ccl}
    a_n(1/2)&=&(-1)^n\displaystyle\prod_{j=1}^n\displaystyle{j+1/2\over2j}= (-1)^n\prod_{j=1}^n\displaystyle{2j+1\over4j},\\
    &=&\displaystyle{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!},\quad n\ge0.
    \end{array}
    \end{equation}

    Substituting this into \eqref{eq:3.6.16} yields

    \begin{eqnarray*}
    y_1=x^{1/2}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}x^n.
    \end{eqnarray*}

    To obtain \(y_2\) in \eqref{eq:3.6.17}, we must compute \(a_n'(1/2)\) for \(n=1\), \(2\), \(\dots\). We'll do this by logarithmic differentiation. From \eqref{eq:3.6.18},

    \begin{eqnarray*}
    |a_n(r)|=\prod_{j=1}^n{|j+r|\over|2j+2r-1|},\quad n\ge1.
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    \ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r|-\ln|2j+2r-1|\right).
    \end{eqnarray*}

    Differentiating with respect to \(r\) yields

    \begin{eqnarray*}
    {a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right).
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right).
    \end{eqnarray*}

    Setting \(r=1/2\) here and recalling \eqref{eq:3.6.19} yields

    \begin{equation} \label{eq:3.6.20}
    a'_n(1/2)={(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}\left(\sum_{j=1}^n{1\over j+1/2}-\sum_{j=1}^n{1\over j}\right).
    \end{equation}

    Since

    \begin{eqnarray*}
    {1\over j+1/2}-{1\over j}={j-j-1/2\over j(j+1/2)}=-{1\over j(2j+1)},
    \end{eqnarray*}

    \eqref{eq:3.6.20} can be rewritten as

    \begin{eqnarray*}
    a'_n(1/2)=-{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \sum_{j=1}^n{1\over j(2j+1)}.
    \end{eqnarray*}

    Therefore, from \eqref{eq:3.6.17},

    \begin{eqnarray*}
    y_2=y_1\ln x-x^{1/2}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \left(\sum_{j=1}^n{1\over j(2j+1)}\right)x^n.
    \end{eqnarray*}

    Example \(\PageIndex{3}\)

    Find a fundamental set of Frobenius solutions of

    \begin{equation} \label{eq:3.6.21}
    x^2(2-x^2)y''-2x(1+2x^2)y'+(2-2x^2)y=0.
    \end{equation}

    Give explicit formulas for the coefficients in the solutions.

    Answer

    For \eqref{eq:3.6.21}, the polynomials defined in Theorem \((3.6.1)\) are

    \begin{eqnarray*}
    \begin{array}{ccccc}
    p_0(r)&=&2r(r-1)-2r+2&=&2(r-1)^2,\\
    p_1(r)&=&0,\\
    p_2(r)&=&-r(r-1)-4r-2&=&-(r+1)(r+2).
    \end{array}
    \end{eqnarray*}

    As in Section \(3.5\), since \(p_1\equiv0\), the recurrence formulas of Theorem \((3.6.1)\) imply that \(a_n(r)=0\) if \(n\) is odd, and

    \begin{eqnarray*}
    \begin{array}{ccl}
    a_0(r)&=&1,\\
    a_{2m}(r)&=&-\displaystyle{p_2(2m+r-2)\over p_0(2m+r)}a_{2m-2}(r)\\
    &=&\displaystyle{(2m+r-1)(2m+r)\over2(2m+r-1)^2}a_{2m-2}(r)\\
    &=&\displaystyle{2m+r\over2(2m+r-1)}a_{2m-2}(r),\quad m\ge1.
    \end{array}
    \end{eqnarray*}

    Since \(r_1=1\) is a repeated root of the indicial polynomial \(p_0\), Theorem \((3.6.2)\) implies that

    \begin{equation} \label{eq:3.6.22}
    y_1=x\sum_{m=0}^\infty a_{2m}(1)x^{2m}
    \end{equation}

    and

    \begin{equation} \label{eq:3.6.23}
    y_2=y_1\ln x+x\sum_{m=1}^\infty a'_{2m}(1)x^{2m}
    \end{equation}

    form a fundamental set of Frobenius solutions of \eqref{eq:3.6.21}. We leave it to you to show that

    \begin{equation} \label{eq:3.6.24}
    a_{2m}(r)={1\over2^m}\prod_{j=1}^m{2j+r\over2j+r-1}.
    \end{equation}

    Setting \(r=1\) yields

    \begin{equation} \label{eq:3.6.25}
    a_{2m}(1)={1\over2^m}\prod_{j=1}^m{2j+1\over2j} ={\prod_{j=1}^m(2j+1)\over4^mm!},
    \end{equation}

    and substituting this into \eqref{eq:3.6.22} yields

    \begin{eqnarray*}
    y_1=x\sum_{m=0}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!}x^{2m}.
    \end{eqnarray*}

    To obtain \(y_2\) in \eqref{eq:3.6.23}, we must compute \(a_{2m}'(1)\) for \(m=1\), \(2\), \(\dots\). Again we use logarithmic differentiation. From \eqref{eq:3.6.24},

    \begin{eqnarray*}
    |a_{2m}(r)|={1\over2^m}\prod_{j=1}^m{|2j+r|\over|2j+r-1|}.
    \end{eqnarray*}

    Taking logarithms yields

    \begin{eqnarray*}
    \ln |a_{2m}(r)|=-m\ln2+ \sum^m_{j=1} \left(\ln |2j+r|-\ln|2j+r-1|\right).
    \end{eqnarray*}

    Differentiating with respect to \(r\) yields

    \begin{eqnarray*}
    {a'_{2m}(r)\over a_{2m}(r)}=\sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    a'_{2m}(r)=a_{2m}(r) \sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).
    \end{eqnarray*}

    Setting \(r=1\) and recalling \eqref{eq:3.6.25} yields

    \begin{equation} \label{eq:3.6.26}
    a'_{2m}(1)=\displaystyle{{\prod_{j=1}^m(2j+1)\over4^mm!} \sum_{j=1}^m\left({1\over2j+1}-{1\over2j}\right)}.
    \end{equation}

    Since

    \begin{eqnarray*}
    {1\over2j+1}-{1\over2j}=-{1\over2j(2j+1)},
    \end{eqnarray*}

    \eqref{eq:3.6.26} can be rewritten as

    \begin{eqnarray*}
    a_{2m}'(1) = -\displaystyle{{\prod_{j=1}^m(2j+1)\over2 \cdot 4^mm!} \sum_{j=1}^m \left({1 \over (2j+1)}\right)}.
    \end{eqnarray*}

    Substituting this into \eqref{eq:3.6.23} yields

    \begin{eqnarray*}
    y_2=y_1\ln x-{x\over2}\sum_{m=1}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!} \left(\sum_{j=1}^m{1\over j(2j+1)}\right)x^{2m}.
    \end{eqnarray*}

    If the solution \(y_1=y(x,r_1)\) of \(Ly=0\) reduces to a finite sum, then there's a difficulty in using logarithmic differentiation to obtain the coefficients \(\{a_n'(r_1)\}\) in the second solution. The next example illustrates this difficulty and shows how to overcome it.

    Example \(\PageIndex{4}\)

    Find a fundamental set of Frobenius solutions of

    \begin{equation} \label{eq:3.6.27}
    x^2y''-x(5-x)y'+(9-4x)y=0.
    \end{equation}

    Give explicit formulas for the coefficients in the solutions.

    Answer

    For \eqref{eq:3.6.27} the polynomials defined in Theorem \((3.6.1)\) are

    \begin{eqnarray*}
    \begin{array}{ccccc}
    p_0(r)&=&r(r-1)-5r+9&=&(r-3)^2,\\
    p_1(r)&=&r-4,\\
    p_2(r)&=&0.
    \end{array}
    \end{eqnarray*}

    Since \(r_1=3\) is a repeated zero of the indicial polynomial \(p_0\), Theorem \((3.6.2)\) implies that

    \begin{equation} \label{eq:3.6.28}
    y_1=x^3\sum_{n=0}^\infty a_n(3)x^n
    \end{equation}

    and

    \begin{equation} \label{eq:3.6.29}
    y_2=y_1\ln x+x^3\sum_{n=1}^\infty a_n'(3)x^n
    \end{equation}

    are linearly independent Frobenius solutions of \eqref{eq:3.6.27}. To find the coefficients in \eqref{eq:3.6.28} we use the recurrence formulas

    \begin{eqnarray*}
    \begin{array}{ccl}
    a_0(r)&=&1,\\
    a_n(r)&=&-\displaystyle{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r)\\
    &=&-\displaystyle{n+r-5\over(n+r-3)^2}a_{n-1}(r),\quad n\ge1.
    \end{array}
    \end{eqnarray*}

    We leave it to you to show that

    \begin{equation} \label{eq:3.6.30}
    a_n(r)=(-1)^n\prod_{j=1}^n{j+r-5\over(j+r-3)^2}.
    \end{equation}

    Setting \(r=3\) here yields

    \begin{eqnarray*}
    a_n(3)=(-1)^n\prod_{j=1}^n{j-2\over j^2},
    \end{eqnarray*}

    so \(a_1(3)=1\) and \(a_n(3)=0\) if \(n\ge2\). Substituting these coefficients into \eqref{eq:3.6.28} yields

    \begin{eqnarray*}
    y_1=x^3(1+x).
    \end{eqnarray*}

    To obtain \(y_2\) in \eqref{eq:3.6.29} we must compute \(a_n'(3)\) for \(n=1\), \(2\), \(\dots\). Let's first try logarithmic differentiation. From \eqref{eq:3.6.30},

    \begin{eqnarray*}
    |a_n(r)|=\prod_{j=1}^n{|j+r-5|\over|j+r-3|^2},\quad n\ge1,
    \end{eqnarray*}

    so

    \begin{eqnarray*}
    \ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r-5|-2\ln|j+r-3|\right).
    \end{eqnarray*}

    Differentiating with respect to \(r\) yields

    \begin{eqnarray*}
    {a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).
    \end{eqnarray*}

    Therefore

    \begin{equation} \label{eq:3.6.31}
    a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).
    \end{equation}

    However, we can't simply set \(r=3\) here if \(n\ge2\), since the bracketed expression in the sum corresponding to \(j=2\) contains the term \(1/(r-3)\). In fact, since \(a_n(3)=0\) for \(n\ge2\), the formula \eqref{eq:3.6.31} for \(a_n'(r)\) is actually an indeterminate form at \(r=3\).

    We overcome this difficulty as follows. From \eqref{eq:3.6.30} with \(n=1\),

    \begin{eqnarray*}
    a_1(r)=-{r-4\over (r-2)^2}.
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    a_1'(r)={r-6\over(r-2)^3},
    \end{eqnarray*}

    so

    \begin{equation} \label{eq:3.6.32}
    a_1'(3)=-3.
    \end{equation}

    From \eqref{eq:3.6.30} with \(n\ge2\),

    \begin{eqnarray*}
    a_n(r)=(-1)^n (r-4)(r-3)\,{\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2} =(r-3)c_n(r),
    \end{eqnarray*}

    where

    \begin{eqnarray*}
    c_n(r)=(-1)^n(r-4)\, {\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2},\quad n\ge2.
    \end{eqnarray*}

    Therefore

    \begin{eqnarray*}
    a_n'(r)=c_n(r)+(r-3)c_n'(r),\quad n\ge2,
    \end{eqnarray*}

    which implies that \(a_n'(3)=c_n(3)\) if \(n\ge3\). We leave it to you to verify that

    \begin{eqnarray*}
    a_n'(3)=c_n(3)={(-1)^{n+1}\over n(n-1)n!},\quad n\ge2.
    \end{eqnarray*}

    Substituting this and \eqref{eq:3.6.32} into \eqref{eq:3.6.29} yields

    \begin{eqnarray*}
    y_2=x^3(1+x)\ln x-3x^4-x^3\displaystyle{\sum_{n=2}^\infty {(-1)^n\over n(n-1)n!}x^n}.
    \end{eqnarray*}