# 7.6: The Method of Frobenius II


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.

## Theorem 7.7.1

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}.

## Theorem 7.7.2

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.

## Example 7.7.1

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.

## Example 7.7.2

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},

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

## Example 7.7.3

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.

## Example 7.7.4

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

$y_2=x^3(1+x)\ln x-3x^4-x^3{\sum_{n=2}^\infty {(-1)^n\over n(n-1)n!}x^n}.\nonumber$

This page titled 7.6: The Method of Frobenius II is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.