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3.7: The Method of Frobenius III

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In Sections $$3.5$$ and $$3.6$$ we discussed methods for finding Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated root or distinct real roots that don't differ by an integer. In this section we consider the case where the indicial equation has distinct real roots that differ by an integer. We'll limit our discussion to equations that can be written as

\label{eq:3.7.1}
x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y' +(\gamma_0+\gamma_1x)y=0

or

\begin{eqnarray*}
x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y' +(\gamma_0+\gamma_2x^2)y=0,
\end{eqnarray*}

where the roots of the indicial equation differ by a positive integer.

We begin with a theorem that provides a fundamental set of solutions of equations of the form \eqref{eq:3.7.1}.

Theorem $$\PageIndex{1}$$

Let

\begin{eqnarray*}
Ly= x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y' +(\gamma_0+\gamma_1x)y,
\end{eqnarray*}

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.
\end{eqnarray*}

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

\label{eq:3.7.2}
\begin{array}{rcl}
a_0(r)&=&1,\\
\end{array}

Let $$r_1$$ and $$r_2$$ be the roots of the indicial equation $$p_0(r)=0,$$ and suppose $$r_1=r_2+k,$$ where $$k$$ is a positive integer. Then

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

is a Frobenius solution of $$Ly=0$$. Moreover, if we define

\label{eq:3.7.3}
\begin{array}{rcl}
a_0(r_2)&=&1,\\
\end{array}

and

\label{eq:3.7.4}
C=-{p_1(r_1-1)\over k\alpha_0}a_{k-1}(r_2),

then

\label{eq:3.7.5}
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)

is also a solution of $$Ly=0,$$ and $$\{y_1,y_2\}$$ is a fundamental set of solutions.

Proof

Theorem $$(3.5.3)$$ implies that $$Ly_1=0$$. We'll now show that $$Ly_2=0$$. Since $$L$$ is a linear operator, this is equivalent to showing that

\label{eq:3.7.6}
L\left( x^{r_2}\sum_{n=0}^{k-1}a_n(r_2)x^n\right)+CL\left(y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\right)=0.

To verify this, we'll show that

\label{eq:3.7.7}
L\left(x^{r_2}\sum_{n=0}^{k-1} a_n(r_2)x^n\right)=p_1(r_1-1)a_{k-1}(r_2)x^{r_1}

and

\label{eq:3.7.8}
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}.

This will imply that $$Ly_2=0$$, since substituting \eqref{eq:3.7.7} and \eqref{eq:3.7.8} into \eqref{eq:3.7.6} and using \eqref{eq:3.7.4} yields

\begin{eqnarray*}
Ly_2&=&\left[p_1(r_1-1)a_{k-1}(r_2)+Ck\alpha_0\right]x^{r_1}\\
&=&\left[p_1(r_1-1)a_{k-1}(r_2)-p_1(r_1-1)a_{k-1}(r_2)\right]x^{r_1}=0.
\end{eqnarray*}

We'll prove \eqref{eq:3.7.8} first. From Theorem $$(3.6.1)$$,

\begin{eqnarray*}
L\left(y(x,r)\ln x+x^r\sum_{n=1}^\infty a_n'(r)x^n\right)=p_0'(r)x^r+x^rp_0(r)\ln x.
\end{eqnarray*}

Setting $$r=r_1$$ and recalling that $$p_0(r_1)=0$$ and $$y_1=y(x,r_1)$$ yields

\label{eq:3.7.9}
L\left(y_1\ln x+x^{r_1}\sum_ {n=1}^\infty a_n'(r_1)x^n\right)=p_0'(r_1)x^{r_1}.

Since $$r_1$$ and $$r_2$$ are the roots of the indicial equation, the indicial polynomial can be written as

\begin{eqnarray*}
p_0(r)=\alpha_0(r-r_1)(r-r_2)=\alpha_0\left[r^2-(r_1+r_2)r+r_1r_2\right].
\end{eqnarray*}

Differentiating this yields

\begin{eqnarray*}
p_0'(r)=\alpha_0(2r-r_1-r_2).
\end{eqnarray*}

Therefore $$p_0'(r_1)=\alpha_0(r_1-r_2)=k\alpha_0$$, so \eqref{eq:3.7.9} implies \eqref{eq:3.7.8}.

Before proving \eqref{eq:3.7.7}, we first note $$a_n(r_2)$$ is well defined by \eqref{eq:3.7.3} for $$1\le n\le k-1$$, since $$p_0(n+r_2)\ne0$$ for these values of $$n$$. However, we can't define $$a_n(r_2)$$ for $$n\ge k$$ with \eqref{eq:3.7.3}, since $$p_0(k+r_2)=p_0(r_1)=0$$. For convenience, we define $$a_n(r_2)=0$$ for $$n\ge k$$. Then, from Theorem $$(3.5.1)$$,

\label{eq:3.7.10}
L\left(x^{r_2}\sum_{n=0}^{k-1} a_n(r_2)x^n\right)= L\left(x^{r_2}\sum_{n=0}^\infty a_n(r_2)x^n\right)= x^{r_2}\sum_{n=0}^\infty b_nx^n,

where $$b_0=p_0(r_2)=0$$ and

\begin{eqnarray*}
\end{eqnarray*}

If $$1\le n\le k-1$$, then \eqref{eq:3.7.3} implies that $$b_n=0$$. If $$n\ge k+1$$, then $$b_n=0$$ because $$a_{n-1}(r_2)=a_n(r_2)=0$$. Therefore \eqref{eq:3.7.10} reduces to

\begin{eqnarray*}
L\left(x^{r_2}\sum_{n=0}^{k-1} a_n(r_2)x^n\right)= \left[p_0(k+r_2)a_k(r_2)+p_1(k+r_2-1)a_{k-1}(r_2) \right]x^{k+r_2}.
\end{eqnarray*}

Since $$a_k(r_2)=0$$ and $$k+r_2=r_1$$, this implies \eqref{eq:3.7.7}.

We leave the proof that $$\{y_1,y_2\}$$ is a fundamental set as an exercise (Exercise $$(3.7.41)$$).

Example $$\PageIndex{1}$$

Find a fundamental set of Frobenius solutions of

\begin{eqnarray*}
2x^2(2+x)y''-x(4-7x)y'-(5-3x)y=0.
\end{eqnarray*}

Give explicit formulas for the coefficients in the solutions.

For the given equation, the polynomials defined in Theorem $$(3.7.1)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&4r(r-1)-4r-5&=&(2r+1)(2r-5),\\
p_1(r)&=&2r(r-1)+7r+3&=&(r+1)(2r+3).
\end{array}
\end{eqnarray*}

The roots of the indicial equation are $$r_1=5/2$$ and $$r_2=-1/2$$, so $$k=r_1-r_2=3$$. Therefore Theorem $$(3.7.1)$$ implies that

\label{eq:3.7.11}
y_1=x^{5/2}\sum_{n=0}^\infty a_n(5/2)x^n

and

\label{eq:3.7.12}
y_2=x^{-1/2}\sum_{n=0}^2a_n(-1/2)+C\left(y_1\ln x+x^{5/2}\sum_{n=1}^\infty a_n'(5/2)x^n\right)

(with $$C$$ as in \eqref{eq:3.7.4}) form a fundamental set of solutions of $$Ly=0$$. The recurrence formula \eqref{eq:3.7.2} is

\label{eq:3.7.13}
\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)(2n+2r-5)}a_{n-1}(r),\\
&=&-\displaystyle{n+r\over2n+2r-5}a_{n-1}(r),\,n\ge1,
\end{array}

which implies that

\label{eq:3.7.14}
a_n(r)=(-1)^n\prod_{j=1}^n{j+r\over2j+2r-5},\,n\ge0.

Therefore

\label{eq:3.7.15}
a_n(5/2)={(-1)^n\prod_{j=1}^n(2j+5)\over4^nn!}.

Substituting this into \eqref{eq:3.7.11} yields

\begin{eqnarray*}
y_1=x^{5/2}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j+5)\over4^nn!}x^n.
\end{eqnarray*}

To compute the coefficients $$a_0(-1/2),a_1(-1/2)$$ and $$a_2(-1/2)$$ in $$y_2$$, we set $$r=-1/2$$ in \eqref{eq:3.7.13} and apply the resulting recurrence formula for $$n=1$$, $$2$$; thus,

\begin{eqnarray*}
a_0(-1/2)&=&1,\\
a_n(-1/2)&=&-{2n-1\over4(n-3)}a_{n-1}(-1/2),\,n=1,2.
\end{eqnarray*}

The last formula yields

\begin{eqnarray*}
\end{eqnarray*}

Substituting $$r_1=5/2,r_2=-1/2,k=3$$, and $$\alpha_0=4$$ into \eqref{eq:3.7.4} yields $$C=-15/128$$. Therefore, from \eqref{eq:3.7.12},

\label{eq:3.7.16}
y_2=x^{-1/2}\left(1+{1\over8}x+{3\over32}x^2\right) -{15\over128} \left(y_1\ln x+x^{5/2}\sum_{n=1}^\infty a_n'(5/2)x^n\right).

We use logarithmic differentiation to obtain obtain $$a'_n(r)$$. From \eqref{eq:3.7.14},

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

Therefore

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

Therefore

\begin{eqnarray*}
a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-5}\right).
\end{eqnarray*}

Setting $$r=5/2$$ here and recalling \eqref{eq:3.7.15} yields

\label{eq:3.7.17}
a_n'(5/2)={(-1)^n\prod_{j=1}^n(2j+5)\over4^nn!}\sum_{j=1}^n \left({1\over j+5/2}-{1\over j}\right).

Since

\begin{eqnarray*}
{1\over j+5/2}-{1\over j}=-{5\over j(2j+5)},
\end{eqnarray*}

we can rewrite \eqref{eq:3.7.17} as

\begin{eqnarray*}
a_n'(5/2)=-5{(-1)^n\prod_{j=1}^n(2j+5)\over4^nn!} \left(\sum_{j=1}^n{1\over j(2j+5)}\right).
\end{eqnarray*}

Substituting this into \eqref{eq:3.7.16} yields

\begin{eqnarray*}
y_2&=&x^{-1/2}\left(1+{1\over8}x+{3\over32}x^2\right)-{15\over128} y_1\ln x\\
&&\,+{75\over128} x^{5/2}\sum_{n=1}^\infty {(-1)^n\prod_{j=1}^n(2j+5)\over4^nn!} \left(\sum_{j=1}^n{1\over j(2j+5)}\right)x^n.
\end{eqnarray*}

If $$C=0$$ in \eqref{eq:3.7.4}, there's no need to compute

\begin{eqnarray*}
y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n
\end{eqnarray*}

in the formula \eqref{eq:3.7.5} for $$y_2$$. Therefore it's best to compute $$C$$ before computing $$\{a_n'(r_1)\}_{n=1}^\infty$$. This is illustrated in the next example. (See also Exercises $$(3.7E.44)$$ and $$(3.7E.45)$$.)

Example $$\PageIndex{2}$$

Find a fundamental set of Frobenius solutions of

\begin{eqnarray*}
x^2(1-2x)y''+x(8-9x)y'+(6-3x)y=0.
\end{eqnarray*}

Give explicit formulas for the coefficients in the solutions.

For the given equation, the polynomials defined in Theorem $$(3.7.1)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&r(r-1)+8r+6&=&(r+1)(r+6)\\
p_1(r)&=&-2r(r-1)-9r-3&=&-(r+3)(2r+1).
\end{array}
\end{eqnarray*}

The roots of the indicial equation are $$r_1=-1$$ and $$r_2=-6$$, so $$k=r_1-r_2=5$$. Therefore Theorem $$(3.7.1)$$ implies that

\label{eq:3.7.18}
y_1=x^{-1}\sum_{n=0}^\infty a_n(-1)x^n

and

\label{eq:3.7.19}
y_2=x^{-6}\sum_{n=0}^4a_n(-6)+C\left(y_1\ln
x+x^{-1}\sum_{n=1}^\infty a_n'(-1)x^n\right)

(with $$C$$ as in \eqref{eq:3.7.4}) form a fundamental set of solutions of $$Ly=0$$. The recurrence formula \eqref{eq:3.7.2} is

\label{eq:3.7.20}
\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+2)(2n+2r-1)\over(n+r+1)(n+r+6)}a_{n-1}(r),\,n\ge1,
\end{array}

which implies that

\label{eq:3.7.21}
\begin{array}{ccl}
a_n(r)&=&\displaystyle{\prod_{j=1}^n{(j+r+2)(2j+2r-1)\over(j+r+1)(j+r+6)}}\\
&=&\displaystyle{\left(\prod_{j=1}^n{j+r+2\over j+r+1}\right) \left(\prod_{j=1}^n{2j+2r-1\over j+r+6}\right)}.
\end{array}

Since

\begin{eqnarray*}
\prod_{j=1}^n{j+r+2\over j+r+1}={(r+3)(r+4)\cdots(n+r+2)\over (r+2)(r+3)\cdots(n+r+1)}={n+r+2\over r+2}
\end{eqnarray*}

because of cancellations, \eqref{eq:3.7.21} simplifies to

\begin{eqnarray*}
a_n(r)={n+r+2\over r+2}\prod_{j=1}^n{2j+2r-1\over j+r+6}.
\end{eqnarray*}

Therefore

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

Substituting this into \eqref{eq:3.7.18} yields

\begin{eqnarray*}
y_1=x^{-1}\sum_{n=0}^\infty (n+1)\left(\prod_{j=1}^n{2j-3\over j+5}\right) x^n.
\end{eqnarray*}

To compute the coefficients $$a_0(-6), \dots ,a_4(-6)$$ in $$y_2$$, we set $$r=-6$$ in \eqref{eq:3.7.20} and apply the resulting recurrence formula for $$n=1$$, $$2$$, $$3$$, $$4$$; thus,

\begin{eqnarray*}
a_0(-6)&=&1,\\
a_n(-6)&=&\displaystyle{(n-4)(2n-13)\over n(n-5)}a_{n-1}(-6),\,n=1,2,3,4.
\end{eqnarray*}

The last formula yields

\begin{eqnarray*}
a_1(-6)=-{33\over 4},\,a_2(-6)={99\over4},\,a_3(-6)=-{231\over8},\,a_4(-6)=0.
\end{eqnarray*}

Since $$a_4(-6)=0$$, \eqref{eq:3.7.4} implies that the constant $$C$$ in \eqref{eq:3.7.19} is zero. Therefore \eqref{eq:3.7.19} reduces to

\begin{eqnarray*}
y_2=x^{-6}\left(1-{33\over4}x+{99\over4}x^2-{231\over8}x^3\right).
\end{eqnarray*}

We now consider equations of the form

\begin{eqnarray*}
x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y' +(\gamma_0+\gamma_2x^2)y=0,
\end{eqnarray*}

where the roots of the indicial equation are real and differ by an even integer. The case where the roots are real and differ by an odd integer can be handled by the method discussed in Exercise $$(3.7E.46)$$.

The proof of the next theorem is similar to the proof of Theorem $$(3.7.1)$$ (Exercise $$(3.7E.43)$$).

Theorem $$\PageIndex{2}$$

Let

\begin{eqnarray*}
Ly= x^2(\alpha_0+\alpha_2x^2)y''+x(\beta_0+\beta_2x^2)y' +(\gamma_0+\gamma_2x^2)y,
\end{eqnarray*}

where $$\alpha_0\ne0,$$ and define

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

Suppose $$r$$ is a real number such that $$p_0(2m+r)$$ is nonzero for all positive integers $$m,$$ and define

\label{eq:3.7.22}
\begin{array}{rcl}
a_0(r)&=&1,\\
\end{array}

Let $$r_1$$ and $$r_2$$ be the roots of the indicial equation $$p_0(r)=0,$$ and suppose $$r_1=r_2+2k,$$ where $$k$$ is a positive integer. Then

\begin{eqnarray*}
y_1=x^{r_1}\sum_{m=0}^\infty a_{2m}(r_1)x^{2m}
\end{eqnarray*}

is a Frobenius solution of $$Ly=0$$. Moreover, if we define

\begin{eqnarray*}
a_0(r_2)&=&1,\\
\end{eqnarray*}

and

\label{eq:3.7.23}
C=-{p_2(r_1-2)\over2k\alpha_0}a_{2k-2}(r_2),

then

\label{eq:3.7.24}
y_2=x^{r_2}\sum_{m=0}^{k-1}a_{2m}(r_2)x^{2m}+C\left(y_1\ln x+x^{r_1}\sum_{m=1}^\infty a_{2m}'(r_1)x^{2m}\right)

is also a solution of $$Ly=0,$$ and $$\{y_1,y_2\}$$ is a fundamental set of solutions.

Proof

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Example $$\PageIndex{3}$$

Find a fundamental set of Frobenius solutions of

\begin{eqnarray*}
x^2(1+x^2)y''+x(3+10x^2)y'-(15-14x^2)y=0.
\end{eqnarray*}

Give explicit formulas for the coefficients in the solutions.

For the given equation, the polynomials defined in Theorem $$(3.7.2)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&r(r-1)+3r-15&=&(r-3)(r+5)\\
p_2(r)&=&r(r-1)+10r+14&=&(r+2)(r+7).
\end{array}
\end{eqnarray*}

The roots of the indicial equation are $$r_1=3$$ and $$r_2=-5$$, so $$k=(r_1-r_2)/2=4$$. Therefore Theorem $$(3.7.2)$$ implies that

\label{eq:3.7.25}
y_1=x^3\sum_{m=0}^\infty a_{2m}(3)x^{2m}

and

\begin{eqnarray*}
y_2=x^{-5}\sum_{m=0}^3 a_{2m}(-5)x^{2m}+C \left(y_1\ln x+x^3\sum_{m=1}^\infty a_{2m}'(3)x^{2m}\right)
\end{eqnarray*}

(with $$C$$ as in \eqref{eq:3.7.23}) form a fundamental set of solutions of $$Ly=0$$. The recurrence formula \eqref{eq:3.7.22} is

\label{eq:3.7.26}
\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)(2m+r+5)\over(2m+r-3)(2m+r+5)}a_{2m-2}(r)\\
&=&-\displaystyle{2m+r\over2m+r-3}a_{2m-2}(r),\,m\ge1,
\end{array}

which implies that

\label{eq:3.7.27}
a_{2m}(r)=(-1)^m\prod_{j=1}^m{2j+r\over2j+r-3},\,m\ge0.

Therefore

\label{eq:3.7.28}
a_{2m}(3)={(-1)^m\prod_{j=1}^m(2j+3)\over2^mm!}.

Substituting this into \eqref{eq:3.7.25} yields

\begin{eqnarray*}
y_1=x^3\sum_{m=0}^\infty {(-1)^m\prod_{j=1}^m(2j+3)\over2^mm!}x^{2m}.
\end{eqnarray*}

To compute the coefficients $$a_2(-5)$$, $$a_4(-5)$$, and $$a_6(-5)$$ in $$y_2$$, we set $$r=-5$$ in \eqref{eq:3.7.26} and apply the resulting recurrence formula for $$m=1$$, $$2$$, $$3$$; thus,

\begin{eqnarray*}
a_{2m}(-5)=-{2m-5\over2(m-4)}a_{2m-2}(-5),\,m=1,2,3.
\end{eqnarray*}

This yields

\begin{eqnarray*}
a_2(-5)=-{1\over2},\,a_4(-5)={1\over8},\,a_6(-5)={1\over16}.
\end{eqnarray*}

Substituting $$r_1=3$$, $$r_2=-5$$, $$k=4$$, and $$\alpha_0=1$$ into \eqref{eq:3.7.23} yields $$C=-3/16$$. Therefore, from \eqref{eq:3.7.24},

\label{eq:3.7.29}
y_2=x^{-5} \left(1-{1\over2}x^2+{1\over8}x^4+{1\over16}x^6\right) -{3\over16} \left(y_1\ln x+x^3\sum_{m=1}^\infty a_{2m}'(3)x^{2m}\right).

To obtain $$a_{2m}'(r)$$ we use logarithmic differentiation. From \eqref{eq:3.7.27},

\begin{eqnarray*}
|a_{2m}(r)|=\prod_{j=1}^m{|2j+r|\over|2j+r-3|},\,m\ge1.
\end{eqnarray*}

Therefore

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

Therefore

\begin{eqnarray*}
a'_{2m}(r)=a_{2m}(r) \sum^n_{j=1} \left({1\over 2j+r}-{1\over2j+r-3}\right).
\end{eqnarray*}

Setting $$r=3$$ here and recalling \eqref{eq:3.7.28} yields

\label{eq:3.7.30}
a_{2m}'(3)={(-1)^m\prod_{j=1}^m(2j+3)\over2^mm!}\sum_{j=1}^m \left({1\over2j+3}-{1\over2j}\right).

Since

\begin{eqnarray*}
{1\over2j+3}-{1\over2j}=-{3\over2j(2j+3)},
\end{eqnarray*}

we can rewrite \eqref{eq:3.7.30} as

\begin{eqnarray*}
a_{2m}'(3)=-{3\over2}{(-1)^n\prod_{j=1}^m(2j+3)\over2^mm!} \left(\sum_{j=1}^n{1\over j(2j+3)}\right).
\end{eqnarray*}

Substituting this into \eqref{eq:3.7.29} yields

\begin{eqnarray*}
y_2&=&x^{-5} \left(1-{1\over2}x^2+{1\over8}x^4+{1\over16}x^6\right) -{3\over16}y_1\ln x \\
&&\, +{9\over32} x^3\sum_{m=1}^\infty {(-1)^m\prod_{j=1}^m(2j+3)\over2^mm!}\left(\sum_{j=1}^m{1\over j(2j+3)}\right) x^{2m}.
\end{eqnarray*}

Example $$\PageIndex{4}$$

Find a fundamental set of Frobenius solutions of

\begin{eqnarray*}
x^2(1-2x^2)y''+x(7-13x^2)y'-14x^2y=0.
\end{eqnarray*}

Give explicit formulas for the coefficients in the solutions.

For the given equation, the polynomials defined in Theorem $$(3.7.2)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&r(r-1)+7r&=&r(r+6),\\
p_2(r)&=&-2r(r-1)-13r-14&=&-(r+2)(2r+7).
\end{array}
\end{eqnarray*}

The roots of the indicial equation are $$r_1=0$$ and $$r_2=-6$$, so $$k=(r_1-r_2)/2=3$$. Therefore Theorem $$(3.7.2)$$ implies that

\label{eq:3.7.31}
y_1=\sum_{m=0}^\infty a_{2m}(0)x^{2m},

and

\label{eq:3.7.32}
y_2=x^{-6}\sum_{m=0}^2a_{2m}(-6)x^{2m}+C\left(y_1\ln x+\sum_{m=1}^\infty a_{2m}'(0)x^{2m}\right)

(with $$C$$ as in \eqref{eq:3.7.23}) form a fundamental set of solutions of $$Ly=0$$. The recurrence formulas \eqref{eq:3.7.22} are

\label{eq:3.7.33}
\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)(4m+2r+3)\over(2m+r)(2m+r+6)}a_{2m-2}(r)\\
&=&\displaystyle{4m+2r+3\over2m+r+6}a_{2m-2}(r),\,m\ge1,
\end{array}

which implies that

\begin{eqnarray*}
a_{2m}(r)=\prod_{j=1}^m{4j+2r+3\over2j+r+6}.
\end{eqnarray*}

Setting $$r=0$$ yields

\begin{eqnarray*}
a_{2m}(0)=6{\prod_{j=1}^m(4j+3)\over2^m(m+3)!}.
\end{eqnarray*}

Substituting this into \eqref{eq:3.7.31} yields

\begin{eqnarray*}
y_1=6\sum_{m=0}^\infty {\prod_{j=1}^m(4j+3)\over2^m(m+3)!}x^{2m}.
\end{eqnarray*}

To compute the coefficients $$a_0(-6)$$, $$a_2(-6)$$, and $$a_4(-6)$$ in $$y_2$$, we set $$r=-6$$ in \eqref{eq:3.7.33} and apply the resulting recurrence formula for $$m=1$$, $$2$$; thus,

\begin{eqnarray*}
a_0(-6)&=&1,\\
a_{2m}(-6)&=&\displaystyle{4m-9\over2m}a_{2m-2}(-6),\,m=1,2.
\end{eqnarray*}

The last formula yields

\begin{eqnarray*}
Since $$p_2(-2)=0$$, the constant $$C$$ in \eqref{eq:3.7.23} is zero. Therefore \eqref{eq:3.7.32} reduces to