7.7: The Method of Frobenius II
- Page ID
- 9444
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
\[\label{eq:7.6.1} x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0\]
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 7.5.3 implies that Equation \ref{eq:7.6.1} has one solution of the form
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 7.5.2 provides a way to find a second solution.
Let
\[\label{eq:7.6.2} Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y,\]
where \(\alpha_0\ne0\) and define
\[\begin{align*} p_0(r)&= \alpha_0r(r-1)+\beta_0r+\gamma_0,\\[4pt] p_1(r)&=\alpha_1r(r-1)+\beta_1r+\gamma_1,\\[4pt] p_2(r)&=\alpha_2r(r-1)+\beta_2r+\gamma_2. \end{align*} \nonumber \]
Suppose \(r\) is a real number such that \(p_0(n+r)\) is nonzero for all positive integers \(n\), and define
\[\begin{align*} a_0(r) &= 1,\\ a_1(r) &= -{p_1(r)\over p_0(r+1)},\\[10pt] a_n(r) &= -{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)},\quad n\ge2. \end{align*} \nonumber \]
Then the Frobenius series
\[\label{eq:7.6.3} y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n\]
satisfies
\[\label{eq:7.6.4} Ly(x,r)=p_0(r)x^r.\]
Moreover\(,\)
\[\label{eq:7.6.5} {\partial y\over \partial r}(x,r)=y(x,r)\ln x+x^r\sum_{n=1}^\infty a_n'(r) x^n,\]
and
\[\label{eq:7.6.6} L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.\]
- Proof
-
Theorem 7.5.2 implies Equation \ref{eq:7.6.4}. Differentiating formally with respect to \(r\) in Equation \ref{eq:7.6.3} yields
\[\begin{aligned} {\partial y\over \partial r}(x,r)&={{\partial\over\partial r}(x^r)\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\[10pt] &={x^r\ln x\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\[10pt] &=y(x,r) \ln x + x^r\sum_{n=1}^\infty a_n'(r)x^n,\end{aligned}\nonumber\]
which proves Equation \ref{eq:7.6.5}.
To prove that \(\partial y(x,r)/\partial r\) satisfies Equation \ref{eq:7.6.6}, we view \(y\) in Equation \ref{eq:7.6.2} as a function \(y=y(x,r)\) of two variables, where the prime indicates partial differentiation with respect to \(x\); thus,
\[y'=y'(x,r)={\partial y\over\partial x}(x,r)\quad \text{and} \quad y''=y''(x,r)={\partial^2 y\over\partial x^2}(x,r).\nonumber \]
With this notation we can use Equation \ref{eq:7.6.2} to rewrite Equation \ref{eq:7.6.4} as
\[\label{eq:7.6.7} x^2q_0(x){\partial^2 y\over \partial x^2}(x,r)+xq_1(x){\partial y\over \partial x}(x,r)+q_2(x)y(x,r)=p_0(r)x^r,\]
where
\[\begin{aligned} q_0(x) &= \alpha_0+\alpha_1x+\alpha_2x^2,\\[4pt] q_1(x) &= \beta_0+\beta_1x+\beta_2x^2,\\[4pt] q_2(x) &= \gamma_0+\gamma_1x+\gamma_2x^2.\\[4pt]\end{aligned}\]
Differentiating both sides of Equation \ref{eq:7.6.7} with respect to \(r\) yields
\[x^2q_0(x){\partial^3y\over \partial r\partial x^2}(x,r)+ xq_1(x){\partial^2y\over \partial r\partial x}(x,r)+q_2(x){\partial y\over\partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x. \nonumber\]
By changing the order of differentiation in the first two terms on the left we can rewrite this as
\[x^2q_0(x){\partial^3 y\over \partial x^2\partial r}(x,r) +xq_1(x){\partial^2 y\over \partial x\partial r}(x,r)+q_2(x){\partial y\over \partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x, \nonumber\]
or
\[x^2q_0(x){\partial^2\over \partial x^2} \left({\partial y\over\partial r}(x,r)\right) +xq_1(x){\partial\over\partial r}\left({\partial y\over\partial x}(x,r)\right) +q_2(x){\partial y\over\partial r}(x,r)= p'_0(r)x^r+p_0(r) x^r \ln x, \nonumber\]
which is equivalent to Equation \ref{eq:7.6.6}.
Let \(L\) be as in Theorem 7.7.1 and suppose the indicial equation \(p_0(r)=0\) has a repeated real root \(r_1.\) Then
\[y_1(x)=y(x,r_1)=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n \nonumber\]
and
\[\label{eq:7.6.8} y_2(x)={\partial y\over\partial r}(x,r_1)=y_1(x)\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\]
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
\[p_0(r)=\alpha_0(r-r_1)^2,\nonumber\]
so
\[p_0(n+r_1)=\alpha_0n^2,\nonumber\]
which is nonzero if \(n>0\). Therefore the assumptions of Theorem 7.7.1 hold with \(r=r_1\), and Equation \ref{eq:7.6.4} implies that \(Ly_1=p_0(r_1)x^{r_1}=0\). Since
\[p_0'(r)=2\alpha(r-r_1) \nonumber\]
it follows that \(p_0'(r_1)=0\), so Equation \ref{eq:7.6.6} implies that
\[Ly_2=p_0'(r_1)x^{r_1}+x^{r_1}p_0(r_1)\ln x=0.\nonumber\]
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 7.6.53.
Find a fundamental set of solutions of
\[\label{eq:7.6.9} x^2(1-2x+x^2)y''-x(3+x)y'+(4+x)y=0.\]
Compute just the terms involving \(x^{n+r_1}\), where \(0\le n\le4\) and \(r_1\) is the root of the indicial equation.
Solution
For the given equation, the polynomials defined in Theorem 7.7.1 are
\[\begin{align*} p_0(r) &= r(r-1)-3r+4 \\[4pt] &= (r-2)^2,\\[4pt] p_1(r) &= -2r(r-1)-r+1 \\[4pt] &= -(r-1)(2r+1),\\[4pt] p_2(r) &= r(r-1). \end{align*} \nonumber \]
Since \(r_1=2\) is a repeated root of the indicial polynomial \(p_0\), Theorem 7.7.2 implies that
\[\label{eq:7.6.10} y_1=x^2\sum_{n=0}^\infty a_n(2)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^2\sum_{n=1}^\infty a_n'(2)x^n\]
form a fundamental set of Frobenius solutions of Equation \ref{eq:7.6.9}. To find the coefficients in these series, we use the recurrence formulas from Theorem 7.7.1 :
\[\label{eq:7.6.11} \begin{align*} a_0(r) &= 1,\\ a_1(r) &= -{p_1(r)\over p_0(r+1)} =-{(r-1)(2r+1)\over(r-1)^2} ={2r+1\over r-1},\\ a_n(r) &= -{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)}\\ &= {(n+r-2)\left[(2n+2r-1)a_{n-1}(r) -(n+r-3)a_{n-2}(r)\right]\over(n+r-2)^2}\\ &= {{(2n+2r-1)\over(n+r-2)}a_{n-1}(r)- {(n+r-3)\over(n+r-2)}a_{n-2}(r)}, \quad n\ge2. \end{align*}\]
Differentiating yields
\[\label{eq:7.6.12} \begin{align*} a'_1(r) &= -{3\over (r-1)^2},\\ a'_n(r) &= {{2n+2r-1\over n+r-2}a'_{n-1}(r)-{n+r-3\over n+r-2}a'_{n-2}(r)}\\ &{-{3\over(n+r-2)^2}a_{n-1}(r)-{1\over(n+r-2)^2}a_{n-2}(r)},\quad n\ge2. \end{align*}\]
Setting \(r=2\) in Equation \ref{eq:7.6.11} and Equation \ref{eq:7.6.12} yields
\[\label{eq:7.6.13} \begin{array}{lll} a_0(2) &= 1,\\ a_1(2) &= 5,\\[10pt] a_n(2) &= {{(2n+3)\over n} a_{n-1}(2)-{(n-1)\over n}a_{n-2}(2)},\quad n\ge2 \end{array}\]
and
\[\label{eq:7.6.14} \begin{array}{lll} a_1'(2) &= -3,\\[10pt] a'_n(2) &= {{2n+3\over n}a'_{n-1}(2)-{n-1\over n}a'_{n-2}(2) -{3\over n^2}a_{n-1}(2)-{1\over n^2}a_{n-2}(2)},\quad n\ge2. \end{array}\]
Computing recursively with Equation \ref{eq:7.6.13} and Equation \ref{eq:7.6.14} yields
\[a_0(2)=1,\,a_1(2)=5,\,a_2(2)=17,\,a_3(2)={143\over3},\,a_4(2)={355\over3},\nonumber\]
and
\[a_1'(2)=-3,\,a_2'(2)=-{29\over2},\,a_3'(2)=-{859\over18}, \,a_4'(2)=-{4693\over36}.\nonumber\]
Substituting these coefficients into Equation \ref{eq:7.6.10} yields
\[y_1=x^2\left(1+5x+17x^2+{143\over3}x^3 +{355\over3}x^4+\cdots\right) \nonumber\]
and
\[y_2=y_1 \ln x -x^3\left(3+{29\over2}x+{859\over18}x^2+{4693\over36}x^3 +\cdots\right). \nonumber \]
Since the recurrence formula Equation \ref{eq:7.6.11} involves three terms, it is not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of Equation \ref{eq:7.6.9}. 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 Equation \ref{eq:7.6.1}. In this case, it is often possible to find explicit formulas for the coefficients. The next two examples illustrate this.
Find a fundamental set of Frobenius solutions of
\[\label{eq:7.6.15} 2x^2(2+x)y''+5x^2y'+(1+x)y=0.\]
Give explicit formulas for the coefficients in the solutions.
Solution
For the given equation, the polynomials defined in Theorem 7.7.1 are
\[\begin{array}{ccccc} p_0(r) &= 4r(r-1)+1 &= (2r-1)^2,\\[4pt] p_1(r) &= 2r(r-1)+5r+1 &= (r+1)(2r+1),\\[4pt] p_2(r) &= 0. \end{array}\nonumber \]
Since \(r_1=1/2\) is a repeated zero of the indicial polynomial \(p_0\), Theorem 7.7.2 implies that
\[\label{eq:7.6.16} y_1=x^{1/2}\sum_{n=0}^\infty a_n(1/2)x^n\]
and
\[\label{eq:7.6.17} y_2=y_1\ln x+x^{1/2}\sum_{n=1}^\infty a_n'(1/2)x^n\]
form a fundamental set of Frobenius solutions of Equation \ref{eq:7.6.15}. Since \(p_2\equiv0\), the recurrence formulas in Theorem 7.7.1 reduce to
\[\begin{align*} a_0(r) &= 1,\\ a_n(r) &= -{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r),\\[10pt] &= -{(n+r)(2n+2r-1)\over(2n+2r-1)^2}a_{n-1}(r),\\[10pt] &= -{n+r\over2n+2r-1}a_{n-1}(r),\quad n\ge0. \end{align*} \nonumber \]
We leave it to you to show that
\[\label{eq:7.6.18} a_n(r)=(-1)^n\prod_{j=1}^n{j+r\over2j+2r-1},\quad n\ge0.\]
Setting \(r=1/2\) yields
\[\label{eq:7.6.19} \begin{array}{ccl} a_n(1/2) &= (-1)^n\prod_{j=1}^n{j+1/2\over2j}= (-1)^n\prod_{j=1}^n{2j+1\over4j},\\[10pt] &= {(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!},\quad n\ge0. \end{array}\]
Substituting this into Equation \ref{eq:7.6.16} yields
\[y_1=x^{1/2}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}x^n.\nonumber\]
To obtain \(y_2\) in Equation \ref{eq:7.6.17}, we must compute \(a_n'(1/2)\) for \(n=1\), \(2\),…. We’ll do this by logarithmic differentiation. From Equation \ref{eq:7.6.18},
\[|a_n(r)|=\prod_{j=1}^n{|j+r|\over|2j+2r-1|},\quad n\ge1.\nonumber\]
Therefore
\[\ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r|-\ln|2j+2r-1|\right). \nonumber\]
Differentiating with respect to \(r\) yields
\[{a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right). \nonumber\]
Therefore
\[a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right). \nonumber\]
Setting \(r=1/2\) here and recalling Equation \ref{eq:7.6.19} yields
\[\label{eq:7.6.20} a'_n(1/2)={(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}\left(\sum_{j=1}^n{1\over j+1/2}-\sum_{j=1}^n{1\over j}\right).\]
Since
\[{1\over j+1/2}-{1\over j}={j-j-1/2\over j(j+1/2)}=-{1\over j(2j+1)}, \nonumber\]
Equation \ref{eq:7.6.20} can be rewritten as
\[a'_n(1/2)=-{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \sum_{j=1}^n{1\over j(2j+1)}. \nonumber\]
Therefore, from Equation \ref{eq:7.6.17},
\[y_2=y_1\ln x-x^{1/2}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \left(\sum_{j=1}^n{1\over j(2j+1)}\right)x^n. \nonumber\]
Find a fundamental set of Frobenius solutions of
\[\label{eq:7.6.21} x^2(2-x^2)y''-2x(1+2x^2)y'+(2-2x^2)y=0.\]
Give explicit formulas for the coefficients in the solutions.
Solution
For Equation \ref{eq:7.6.21}, the polynomials defined in Theorem 7.7.1 are
\[\begin{align*} p_0(r) &= 2r(r-1)-2r+2 &= 2(r-1)^2,\\[4pt] p_1(r) &= 0,\\[4pt] p_2(r) &= -r(r-1)-4r-2 &= -(r+1)(r+2). \end{align*} \nonumber \]
As in Section 7.5, since \(p_1\equiv0\), the recurrence formulas of Theorem 7.7.1 imply that \(a_n(r)=0\) if \(n\) is odd, and
\[\begin{align*} a_0(r) &= 1,\\ a_{2m}(r) &= -{p_2(2m+r-2)\over p_0(2m+r)}a_{2m-2}(r)\\[10pt] &= {(2m+r-1)(2m+r)\over2(2m+r-1)^2}a_{2m-2}(r)\\[10pt] &= {2m+r\over2(2m+r-1)}a_{2m-2}(r),\quad m\ge1. \end{align*} \nonumber \]
Since \(r_1=1\) is a repeated root of the indicial polynomial \(p_0\), Theorem 7.7.2 implies that
\[\label{eq:7.6.22} y_1=x\sum_{m=0}^\infty a_{2m}(1)x^{2m}\]
and
\[\label{eq:7.6.23} y_2=y_1\ln x+x\sum_{m=1}^\infty a'_{2m}(1)x^{2m}\]
form a fundamental set of Frobenius solutions of Equation \ref{eq:7.6.21}. We leave it to you to show that
\[\label{eq:7.6.24} a_{2m}(r)={1\over2^m}\prod_{j=1}^m{2j+r\over2j+r-1}.\]
Setting \(r=1\) yields
\[\label{eq:7.6.25} a_{2m}(1)={1\over2^m}\prod_{j=1}^m{2j+1\over2j} ={\prod_{j=1}^m(2j+1)\over4^mm!},\]
and substituting this into Equation \ref{eq:7.6.22} yields
\[y_1=x\sum_{m=0}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!}x^{2m}. \nonumber\]
To obtain \(y_2\) in Equation \ref{eq:7.6.23}, we must compute \(a_{2m}'(1)\) for \(m=1\), \(2\), …. Again we use logarithmic differentiation. From Equation \ref{eq:7.6.24},
\[|a_{2m}(r)|={1\over2^m}\prod_{j=1}^m{|2j+r|\over|2j+r-1|}.\nonumber\]
Taking logarithms yields
\[\ln |a_{2m}(r)|=-m\ln2+ \sum^m_{j=1} \left(\ln |2j+r|-\ln|2j+r-1|\right).\nonumber\]
Differentiating with respect to \(r\) yields
\[{a'_{2m}(r)\over a_{2m}(r)}=\sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).\nonumber\]
Therefore
\[a'_{2m}(r)=a_{2m}(r) \sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).\nonumber\]
Setting \(r=1\) and recalling Equation \ref{eq:7.6.25} yields
\[\label{eq:7.6.26} a'_{2m}(1)={{\prod_{j=1}^m(2j+1)\over4^mm!} \sum_{j=1}^m\left({1\over2j+1}-{1\over2j}\right)}.\]
Since
\[{1\over2j+1}-{1\over2j}=-{1\over2j(2j+1)},\nonumber\]
Equation \ref{eq:7.6.26} can be rewritten as
\[a_{2m}'(1)=-\dfrac{\prod_{j=1}^{m}(2j+1)}{2\cdot 4^{m}m!}\sum_{j=1}^{m}\dfrac{1}{j(2j+1)}\nonumber \]
Substituting this into Equation \ref{eq:7.6.23} yields
\[y_2=y_1\ln x-{x\over2}\sum_{m=1}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!} \left(\sum_{j=1}^m{1\over j(2j+1)}\right)x^{2m}.\nonumber \]
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.
Find a fundamental set of Frobenius solutions of
\[\label{eq:7.6.27} x^2y''-x(5-x)y'+(9-4x)y=0.\]
Give explicit formulas for the coefficients in the solutions.
Solution
For Equation \ref{eq:7.6.27} the polynomials defined in Theorem 7.7.1 are
\[\begin{align*} p_0(r) &= r(r-1)-5r+9 &= (r-3)^2,\\[4pt] p_1(r) &= r-4,\\[4pt] p_2(r) &= 0. \end{align*} \nonumber \]
Since \(r_1=3\) is a repeated zero of the indicial polynomial \(p_0\), Theorem 7.7.2 implies that
\[\label{eq:7.6.28} y_1=x^3\sum_{n=0}^\infty a_n(3)x^n\]
and
\[\label{eq:7.6.29} y_2=y_1\ln x+x^3\sum_{n=1}^\infty a_n'(3)x^n\]
are linearly independent Frobenius solutions of Equation \ref{eq:7.6.27}. To find the coefficients in Equation \ref{eq:7.6.28} we use the recurrence formulas
\[\begin{align*} a_0(r) &= 1,\\ a_n(r) &= -{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r)\\[10pt] &= -{n+r-5\over(n+r-3)^2}a_{n-1}(r),\quad n\ge1. \end{align*} \nonumber \]
We leave it to you to show that
\[\label{eq:7.6.30} a_n(r)=(-1)^n\prod_{j=1}^n{j+r-5\over(j+r-3)^2}.\]
Setting \(r=3\) here yields
\[a_n(3)=(-1)^n\prod_{j=1}^n{j-2\over j^2},\nonumber\]
so \(a_1(3)=1\) and \(a_n(3)=0\) if \(n\ge2\). Substituting these coefficients into Equation \ref{eq:7.6.28} yields
\[y_1=x^3(1+x).\nonumber \]
To obtain \(y_2\) in Equation \ref{eq:7.6.29} we must compute \(a_n'(3)\) for \(n=1\), \(2\), …. Let’s first try logarithmic differentiation. From Equation \ref{eq:7.6.30},
\[|a_n(r)|=\prod_{j=1}^n{|j+r-5|\over|j+r-3|^2},\quad n\ge1,\nonumber\]
so
\[\ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r-5|-2\ln|j+r-3|\right).\nonumber\]
Differentiating with respect to \(r\) yields
\[{a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).\nonumber\]
Therefore
\[\label{eq:7.6.31} a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).\]
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 Equation \ref{eq:7.6.31} for \(a_n'(r)\) is actually an indeterminate form at \(r=3\).
We overcome this difficulty as follows. From Equation \ref{eq:7.6.30} with \(n=1\),
\[a_1(r)=-{r-4\over (r-2)^2}.\nonumber\]
Therefore
\[a_1'(r)={r-6\over(r-2)^3},\nonumber\]
so
\[\label{eq:7.6.32} a_1'(3)=-3.\]
From Equation \ref{eq:7.6.30} with \(n\ge2\),
\[a_n(r)=(-1)^n (r-4)(r-3)\,{\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2} =(r-3)c_n(r),\nonumber\]
where
\[c_n(r)=(-1)^n(r-4)\, {\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2},\quad n\ge2.\nonumber\]
Therefore
\[a_n'(r)=c_n(r)+(r-3)c_n'(r),\quad n\ge2,\nonumber\]
which implies that \(a_n'(3)=c_n(3)\) if \(n\ge3\). We leave it to you to verify that
\[a_n'(3)=c_n(3)={(-1)^{n+1}\over n(n-1)n!},\quad n\ge2.\nonumber\]
Substituting this and Equation \ref{eq:7.6.32} into Equation \ref{eq:7.6.29} yields
\[y_2=x^3(1+x)\ln x-3x^4-x^3{\sum_{n=2}^\infty {(-1)^n\over n(n-1)n!}x^n}.\nonumber\]