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

\begin{equation} 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 \tag{7.6.1}\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 7.5.3 implies that (7.6.1) has one solution of the form

\[ y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n, \]

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 7.6.1

Let

\begin{equation} 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, \tag{7.6.2}\end{equation}

where \(\alpha_0\ne0\) and define

\begin{eqnarray*} p_0(r)&=&\alpha_0r(r-1)+\beta_0r+\gamma_0, \\[5pt] p_1(r)&=&\alpha_1r(r-1)+\beta_1r+\gamma_1, \\[5pt] 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

\[ \begin{array}{ccl} a_0(r)&=&1,\\ a_1(r)&=&-\dst{p_1(r)\over p_0(r+1)},\\[10pt] a_n(r)&=&-\dst{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)},\quad n\ge2. \end{array} \]

Then the Frobenius series

\begin{equation} y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n \tag{7.6.3}\end{equation}

satisfies

\begin{equation} Ly(x,r)=p_0(r)x^r. \tag{7.6.4}\end{equation}

Moreover\(,\)

\begin{equation} {\partial y\over \partial r}(x,r)=y(x,r)\ln x+x^r\sum_{n=1}^\infty a_n'(r) x^n, \tag{7.6.5}\end{equation}

and

\begin{equation} L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x. \tag{7.6.6}\end{equation}

Proof Theorem 7.5.2 implies (7.6.4). Differentiating formally with respect to \(r\) in (7.6.3) yields

\begin{eqnarray*} {\partial y\over \partial r}(x,r)&=&\dst{{\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] &=&\dst{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{eqnarray*}

which proves (7.6.5).

To prove that \(\partial y(x,r)/\partial r\) satisfies (7.6.6), we view \(y\) in (7.6.2) as a function \(y=y(x,r)\) of two variables, where the prime indicates partial differentiation with respect to \(x\); thus,

\[ y'=y'(x,r)={\partial y\over\partial x}(x,r)\mbox{\quad and \quad} y''=y''(x,r)={\partial^2 y\over\partial x^2}(x,r). \]

With this notation we can use (7.6.2) to rewrite (7.6.4) as

\begin{equation} 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, \tag{7.6.7}\end{equation}

where

\begin{eqnarray*} q_0(x)&=&\alpha_0+\alpha_1x+\alpha_2x^2, \\[5pt] q_1(x)&=&\beta_0+\beta_1x+\beta_2x^2, \\[5pt] q_2(x)&=&\gamma_0+\gamma_1x+\gamma_2x^2.\end{eqnarray*}

Differentiating both sides of (7.6.7) with respect to \(r\) yields

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

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

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

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

which is equivalent to (7.6.6).

Theorem 7.6.2

Let \(L\) be as in Theorem 7.6.1 and suppose the indicial equation \(p_0(r)=0\) has a repeated real root \(r_1.\) Then

\[ y_1(x)=y(x,r_1)=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n \]

and

\begin{equation} 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 \tag{7.6.8}\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

\[ p_0(r)=\alpha_0(r-r_1)^2, \]

\[ p_0(n+r_1)=\alpha_0n^2, \]

which is nonzero if \(n>0\). Therefore the assumptions of Theorem 7.6.1 hold with \(r=r_1\), and (7.6.4) implies that \(Ly_1=p_0(r_1)x^{r_1}=0\). Since

\[ p_0'(r)=2\alpha(r-r_1) \]

it follows that \(p_0'(r_1)=0\), so (7.6.6) implies that

\[ Ly_2=p_0'(r_1)x^{r_1}+x^{r_1}p_0(r_1)\ln x=0. \]

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 53).

Example 7.6.1

Find a fundamental set of solutions of

\begin{equation} x^2(1-2x+x^2)y''-x(3+x)y'+(4+x)y=0. \tag{7.6.9}\end{equation}

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 7.6.1 are

\[ \begin{array}{lllll} p_0(r)&=&r(r-1)-3r+4&=&(r-2)^2,\\[5pt] p_1(r)&=&-2r(r-1)-r+1&=&-(r-1)(2r+1),\\[5pt] p_2(r)&=&r(r-1). \end{array} \]

Since \(r_1=2\) is a repeated root of the indicial polynomial \(p_0\), Theorem 7.6.2 implies that

\begin{equation} 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 \tag{7.6.10}\end{equation}

form a fundamental set of Frobenius solutions of (7.6.9). To find the coefficients in these series, we use the recurrence formulas from Theorem 7.6.1:

\begin{equation} \begin{array}{ccl} a_0(r)&=&1,\\ a_1(r)&=&-\dst{p_1(r)\over p_0(r+1)} =-\dst{(r-1)(2r+1)\over(r-1)^2} =\dst{2r+1\over r-1},\\[10pt] a_n(r)&=&-\dst{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)}\\[10pt] &=&\dst{(n+r-2)\left[(2n+2r-1)a_{n-1}(r) -(n+r-3)a_{n-2}(r)\right]\over(n+r-2)^2}\\[10pt] &=&\dst{{(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} \tag{7.6.11}\end{equation}

Differentiating yields

\begin{equation} \begin{array}{ccl} a'_1(r)&=&-\dst{3\over (r-1)^2},\\[10pt] a'_n(r)&=&\dst{{2n+2r-1\over n+r-2}a'_{n-1}(r)-{n+r-3\over n+r-2}a'_{n-2}(r)}\\[10pt] &&\dst{-{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} \tag{7.6.12}\end{equation}

Setting \(r=2\) in (7.6.11) and (7.6.12) yields

\begin{equation} \begin{array}{ccl} a_0(2)&=&1,\\ a_1(2)&=&5,\\[10pt] a_n(2)&=&\dst{{(2n+3)\over n} a_{n-1}(2)-{(n-1)\over n}a_{n-2}(2)},\quad n\ge2 \end{array} \tag{7.6.13}\end{equation}

and

\begin{equation} \begin{array}{ccl} a_1'(2)&=&-3,\\[10pt] a'_n(2)&=&\dst{{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} \tag{7.6.14}\end{equation}

Computing recursively with (7.6.13) and (7.6.14) yields

\[ a_0(2)=1,\,a_1(2)=5,\,a_2(2)=17,\,a_3(2)={143\over3},\,a_4(2)={355\over3}, \]

and

\[ a_1'(2)=-3,\,a_2'(2)=-{29\over2},\,a_3'(2)=-{859\over18}, \,a_4'(2)=-{4693\over36}. \]

Substituting these coefficients into (7.6.10) yields

\[ y_1=x^2\left(1+5x+17x^2+{143\over3}x^3 +{355\over3}x^4+\cdots\right) \]

and

\[ y_2=y_1 \ln x -x^3\left(3+{29\over2}x+{859\over18}x^2+{4693\over36}x^3 +\cdots\right). \bbox \]

Since the recurrence formula (7.6.11) involves three terms, it’s not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of (7.6.9). However, as we saw in the preceding sections, the recurrrence 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 (7.6.1). In this case, it’s often possible to find explicit formulas for the coefficients. The next two examples illustrate this.

Example 7.6.2

Find a fundamental set of Frobenius solutions of

\begin{equation} 2x^2(2+x)y''+5x^2y'+(1+x)y=0. \tag{7.6.15}\end{equation}

Give explicit formulas for the coefficients in the solutions.

Solution For the given equation, the polynomials defined in Theorem 7.6.1 are

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

Since \(r_1=1/2\) is a repeated zero of the indicial polynomial \(p_0\), Theorem 7.6.2 implies that

\begin{equation} y_1=x^{1/2}\sum_{n=0}^\infty a_n(1/2)x^n \tag{7.6.16}\end{equation}

and

\begin{equation} y_2=y_1\ln x+x^{1/2}\sum_{n=1}^\infty a_n'(1/2)x^n \tag{7.6.17}\end{equation}

form a fundamental set of Frobenius solutions of (7.6.15). Since \(p_2\equiv0\), the recurrence formulas in Theorem 7.6.1 reduce to

\[ \begin{array}{ccl} a_0(r)&=&1,\\ a_n(r)&=&-\dst{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r),\\[10pt] &=&-\dst{(n+r)(2n+2r-1)\over(2n+2r-1)^2}a_{n-1}(r),\\[10pt] &=&-\dst{n+r\over2n+2r-1}a_{n-1}(r),\quad n\ge0. \end{array} \]

We leave it to you to show that

\begin{equation} a_n(r)=(-1)^n\prod_{j=1}^n{j+r\over2j+2r-1},\quad n\ge0. \tag{7.6.18}\end{equation}

Setting \(r=1/2\) yields

\begin{equation} \begin{array}{ccl} a_n(1/2)&=&(-1)^n\dst\prod_{j=1}^n\dst{j+1/2\over2j}= (-1)^n\prod_{j=1}^n\dst{2j+1\over4j},\\[10pt] &=&\dst{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!},\quad n\ge0. \end{array} \tag{7.6.19}\end{equation}

Substituting this into (7.6.16) yields

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

To obtain \(y_2\) in (7.6.17), we must compute \(a_n'(1/2)\) for \(n=1\), \(2\),…. We’ll do this by logarithmic differentiation. From (7.6.18),

\[ |a_n(r)|=\prod_{j=1}^n{|j+r|\over|2j+2r-1|},\quad n\ge1. \]

Therefore

\[ \ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r|-\ln|2j+2r-1|\right). \]

Differentiating with respect to \(r\) yields

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

Therefore

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

Setting \(r=1/2\) here and recalling (7.6.19) yields

\begin{equation} 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). \tag{7.6.20}\end{equation}

Since

\[ {1\over j+1/2}-{1\over j}={j-j-1/2\over j(j+1/2)}=-{1\over j(2j+1)}, \]

(7.6.20) can be rewritten as

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

Therefore, from (7.6.17),

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

Example 7.6.3

Find a fundamental set of Frobenius solutions of

\begin{equation} x^2(2-x^2)y''-2x(1+2x^2)y'+(2-2x^2)y=0. \tag{7.6.21}\end{equation}

Give explicit formulas for the coefficients in the solutions.

Solution For (7.6.21), the polynomials defined in Theorem 7.6.1 are

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

As in Section 7.5, since \(p_1\equiv0\), the recurrence formulas of Theorem 7.6.1 imply that \(a_n(r)=0\) if \(n\) is odd, and

\[ \begin{array}{ccl} a_0(r)&=&1,\\ a_{2m}(r)&=&-\dst{p_2(2m+r-2)\over p_0(2m+r)}a_{2m-2}(r)\\[10pt] &=&\dst{(2m+r-1)(2m+r)\over2(2m+r-1)^2}a_{2m-2}(r)\\[10pt] &=&\dst{2m+r\over2(2m+r-1)}a_{2m-2}(r),\quad m\ge1. \end{array} \]

Since \(r_1=1\) is a repeated root of the indicial polynomial \(p_0\), Theorem 7.6.2 implies that

\begin{equation} y_1=x\sum_{m=0}^\infty a_{2m}(1)x^{2m} \tag{7.6.22}\end{equation}

and

\begin{equation} y_2=y_1\ln x+x\sum_{m=1}^\infty a'_{2m}(1)x^{2m} \tag{7.6.23}\end{equation}

form a fundamental set of Frobenius solutions of (7.6.21). We leave it to you to show that

\begin{equation} a_{2m}(r)={1\over2^m}\prod_{j=1}^m{2j+r\over2j+r-1}. \tag{7.6.24}\end{equation}

Setting \(r=1\) yields

\begin{equation} a_{2m}(1)={1\over2^m}\prod_{j=1}^m{2j+1\over2j} ={\prod_{j=1}^m(2j+1)\over4^mm!}, \tag{7.6.25}\end{equation}

and substituting this into (7.6.22) yields

\[ y_1=x\sum_{m=0}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!}x^{2m}. \]

To obtain \(y_2\) in (7.6.23), we must compute \(a_{2m}'(1)\) for \(m=1\), \(2\), …. Again we use logarithmic differentiation. From (7.6.24),

\[ |a_{2m}(r)|={1\over2^m}\prod_{j=1}^m{|2j+r|\over|2j+r-1|}. \]

Taking logarithms yields

\[ \ln |a_{2m}(r)|=-m\ln2+ \sum^m_{j=1} \left(\ln |2j+r|-\ln|2j+r-1|\right). \]

Differentiating with respect to \(r\) yields

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

Therefore

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

Setting \(r=1\) and recalling (7.6.25) yields

\begin{equation} a'_{2m}(1)=\dst{{\prod_{j=1}^m(2j+1)\over4^mm!} \sum_{j=1}^m\left({1\over2j+1}-{1\over2j}\right)}. \tag{7.6.26}\end{equation}

Since

\[ {1\over2j+1}-{1\over2j}=-{1\over2j(2j+1)}, \]

(7.6.26) can be rewritten as

\[ a_{2m}'(1)=-\dst{{\prod_{j=1}^m(2j+1)\over2\cdot4^mm!} \sum_{j=1}^m{1\over j(2j+1)}}. \]

Substituting this into (7.6.23) yields

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

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 7.6.4

Find a fundamental set of Frobenius solutions of

\begin{equation} x^2y''-x(5-x)y'+(9-4x)y=0. \tag{7.6.27}\end{equation}

Give explicit formulas for the coefficients in the solutions.

Solution For (7.6.27) the polynomials defined in Theorem 7.6.1 are

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

Since \(r_1=3\) is a repeated zero of the indicial polynomial \(p_0\), Theorem 7.6.2 implies that

\begin{equation} y_1=x^3\sum_{n=0}^\infty a_n(3)x^n \tag{7.6.28}\end{equation}

and

\begin{equation} y_2=y_1\ln x+x^3\sum_{n=1}^\infty a_n'(3)x^n \tag{7.6.29}\end{equation}

are linearly independent Frobenius solutions of (7.6.27). To find the coefficients in (7.6.28) we use the recurrence formulas

\[ \begin{array}{ccl} a_0(r)&=&1,\\ a_n(r)&=&-\dst{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r)\\[10pt] &=&-\dst{n+r-5\over(n+r-3)^2}a_{n-1}(r),\quad n\ge1. \end{array} \]

We leave it to you to show that

\begin{equation} a_n(r)=(-1)^n\prod_{j=1}^n{j+r-5\over(j+r-3)^2}. \tag{7.6.30}\end{equation}

Setting \(r=3\) here yields

\[ a_n(3)=(-1)^n\prod_{j=1}^n{j-2\over j^2}, \]

so \(a_1(3)=1\) and \(a_n(3)=0\) if \(n\ge2\). Substituting these coefficients into (7.6.28) yields

\[ y_1=x^3(1+x). \]

To obtain \(y_2\) in (7.6.29) we must compute \(a_n'(3)\) for \(n=1\), \(2\), …. Let’s first try logarithmic differentiation. From (7.6.30),

\[ |a_n(r)|=\prod_{j=1}^n{|j+r-5|\over|j+r-3|^2},\quad n\ge1, \]

\[ \ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r-5|-2\ln|j+r-3|\right). \]

Differentiating with respect to \(r\) yields

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

Therefore

\begin{equation} a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right). \tag{7.6.31}\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 (7.6.31) for \(a_n'(r)\) is actually an indeterminate form at \(r=3\).

We overcome this difficulty as follows. From (7.6.30) with \(n=1\),

\[ a_1(r)=-{r-4\over (r-2)^2}. \]

Therefore

\[ a_1'(r)={r-6\over(r-2)^3}, \]

\begin{equation} a_1'(3)=-3. \tag{7.6.32}\end{equation}

From (7.6.30) with \(n\ge2\),

\[ a_n(r)=(-1)^n (r-4)(r-3)\,{\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2} =(r-3)c_n(r), \]

where

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

Therefore

\[ a_n'(r)=c_n(r)+(r-3)c_n'(r),\quad n\ge2, \]

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

\[ a_n'(3)=c_n(3)={(-1)^{n+1}\over n(n-1)n!},\quad n\ge2. \]

Substituting this and (7.6.32) into (7.6.29) yields

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

7.6 Exercises

In Exercises 1–11 find a fundamental set of Frobenius solutions. Compute the terms involving \(x^{n+r_1}\), where \(0\le n\le N\) (\(N\) at least \(7\)) and \(r_1\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(N>7\).

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

Show answer

\(\dst{y_1=x\left(1-x+{3\over4}x^2-{13\over36}x^3+\cdots\right)}\); \(\dst{y_2=y_1 \ln x + x^2\left(1-x+{65\over108}x^2+ \cdots\right)}\)
C \(x^2(1+x+2x^2)y'+x(3+6x+7x^2)y'+(1+6x-3x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1}\left(1-2x+{9\over2}x^2-{20\over3}x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x+1-{15\over4}x+{133\over18}x^2+\cdots}\)
C \(x^2(1+2x+x^2)y''+x(1+3x+4x^2)y'-x(1-2x)y=0\)

Show answer

\(\dst{y_1=1+x-x^2+{1\over3}x^3+\cdots}\); \(\dst{y_2= y_1\ln x-x\left(3-{1\over2}x-{31\over18}x^2+\cdots\right)}\)
C \(4x^2(1+x+x^2)y''+12x^2(1+x)y'+(1+3x+3x^2)y=0\)

Show answer

\(\dst{y_1=x^{1/2}\left(1-2x+{5\over2}x^2-2x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x+x^{3/2}\left(1-{9\over4}x+{17\over6}x^2+ \cdots\right)}\)
C \(x^2(1+x+x^2)y''-x(1-4x-2x^2)y'+y=0\)

Show answer

\(\dst{y_1=x\left(1-4x+{19\over2}x^2-{49\over3}x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x+x^2\left(3-{43\over4}x+{208\over9}x^2+\cdots \right)}\)
C \(9x^2y''+3x(5+3x-2x^2)y'+(1+12x-14x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/3}\left(1-x+{5\over6}x^2-{1\over2}x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x+x^{2/3}\left(1-{11\over12}x+{25\over36}x^2+ \cdots\right)}\)
C \(x^2y''+x(1+x+x^2)y'+x(2-x)y=0\)

Show answer

\(\dst{y_1=1-2x+{7\over4}x^2-{7\over9}x^3+\cdots}\); \(\dst{y_2= y_1\ln x+x\left(3-{15\over4}x+{239\over108}x^2+\cdots \right)}\)
C \(x^2(1+2x)y''+x(5+14x+3x^2)y'+(4+18x+12x^2)y=0\)

Show answer

\(\dst{y_1=x^{-2}\left(1-2x+{5\over2}x^2-3x^3+\cdots\right)}\); \(\dst{y_2=y_1\ln x+{3\over4}-{13\over6}x+\cdots}\)
C \(4x^2y''+2x(4+x+x^2)y'+(1+5x+3x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/2}\left(1-x+{1\over4}x^2+{1\over18}x^3+\cdots\right)}\); \(\dst{y_2=y_1\ln x+x^{1/2}\left({3\over2}-{13\over16}x +{1\over54}x^2+\cdots\right)}\)
C \(16x^2y''+4x(6+x+2x^2)y'+(1+5x+18x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/4}\left(1-{1\over4}x-{7\over32}x^2+{23\over384}x^3 +\cdots\right)}\); \(\dst{y_2= y_1\ln x+x^{3/4}\left({1\over4}+{5\over64}x-{157\over2304}x^2 +\cdots\right)}\)
C \(9x^2(1+x)y''+3x(5+11x-x^2)y'+(1+16x-7x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/3}\left(1-x+{7\over6}x^2-{23\over18}x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x-x^{5/3}\left({1\over12}-{13\over108}x\cdots\right)}\)

In Exercises 12–22 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

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

Show answer

\(\dst{y_1=x^{1/2}\sum_{n=0}^\infty {(-1)^n\over(n!)^2}x^n}\); \(\dst{y_2=y_1\ln x-2x^{1/2}\sum_{n=1}^\infty {(-1)^n\over(n!)^2}\left(\sum_{j=1}^n{1\over j}\right)x^n}\);
\(36x^2(1-2x)y''+24x(1-9x)y'+(1-70x)y=0\)

Show answer

\(\dst{y_1=x^{1/6}\sum_{n=0}^\infty\left(2\over3\right)^n {\prod_{j=1}^n(3j+1)\over n!}x^n}\);

\(\dst{y_2=y_1\ln x-x^{1/6}\sum_{n=1}^\infty \left(2\over3\right)^n {\prod_{j=1}^n(3j+1)\over n!}\left(\sum_{j=1}^n{1\over j(3j+1)}\right)x^n}\)
\(x^2(1+x)y''-x(3-x)y'+4y=0\)

Show answer

\(\dst{y_1=x^2\sum_{n=0}^\infty} (-1)^n(n+1)^2x^n\); \(\dst{y_2=y_1\ln x-2x^2\sum_{n=1}^\infty(-1)^nn(n+1)x^n}\)
\(x^2(1-2x)y''-x(5-4x)y'+(9-4x)y=0\)

Show answer

\(\dst{y_1=x^3\sum_{n=0}^\infty 2^n(n+1)x^n}\); \(\dst{y_2= y_1\ln x-x^3\sum_{n=1}^\infty2^nnx^n}\)
\(25x^2y''+x(15+x)y'+(1+x)y=0\)

Show answer

\(\dst{y_1=x^{1/5}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(5j+1)\over 125^n(n!)^2} x^n}\);

\(\dst{y_2= y_1\ln x-x^{1/5}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(5j+1)\over125^n(n!)^2} \left(\sum_{j=1}^n{5j+2\over j(5j+1)}\right)x^n}\)
\(2x^2(2+x)y''+x^2y'+(1-x)y=0\)

Show answer

\(\dst{y_1=x^{1/2}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j-3)\over4^nn!} x^n}\);

\(\dst{y_2=y_1\ln x+3x^{1/2}\sum_{n=1}^\infty {(-1)^n\prod_{j=1}^n(2j-3)\over4^nn!}\left(\sum_{j=1}^n{1\over j(2j-3)}\right)x^n}\)
\(x^2(9+4x)y''+3xy'+(1+x)y=0\)

Show answer

\(\dst{y_1=x^{1/3}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(6j-7)^2\over 81^n(n!)^2} x^n}\);

\(\dst{y_2=y_1\ln x+14x^{1/3}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(6j-7)^2\over 81^n(n!)^2}\left(\sum_{j=1}^n{1\over j(6j-7)})\right)x^n}\)
\(x^2y''-x(3-2x)y'+(4+3x)y=0\)

Show answer

\(\dst{y_1=x^2\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j+5)\over(n!)^2} x^n}\);

\(\dst{y_2= y_1\ln x-2x^2\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(2j+5)\over(n!)^2} \left(\sum_{j=1}^n{(j+ 5)\over j(2j+ 5)}\right)x^n}\)
\(x^2(1-4x)y''+3x(1-6x)y'+(1-12x)y=0\)

Show answer

\(\dst{y_1={1\over x}\sum_{n=0}^\infty {2^n\prod_{j=1}^n(2j-1)\over n!} x^n}\);

\(\dst{y_2=y_1\ln x+{1\over x}\sum_{n=1}^\infty{2^n\prod_{j=1}^n(2j-1)\over n!} \left(\sum_{j=1}^n{1\over j(2j-1)}\right)x^n}\)
\(x^2(1+2x)y''+x(3+5x)y'+(1-2x)y=0\)

Show answer

\(\dst{y_1={1\over x}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j-5)\over n!}x^n}\);

\(\dst{y_2=y_1\ln x+{5\over x}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(2j-5)\over n!} \left(\sum_{j=1}^n{1\over j(2j-5)}\right)x^n}\)
\(2x^2(1+x)y''-x(6-x)y'+(8-x)y=0\)

Show answer

\(\dst{y_1=x^2\sum_{n=0}^\infty {(-1)^n\prod_{j=1}^n(2j+3)\over2^nn!}x^n}\);

\(\dst{y_2=y_1\ln x-3x^2\sum_{n=0}^\infty {(-1)^n\prod_{j=1}^n(2j+3)\over2^nn!} \left(\sum_{j=1}^n{1\over j(2j+ 3)}\right)x^n}\)

In Exercises 23–27 find a fundamental set of Frobenius solutions. Compute the terms involving \(x^{n+r_1}\), where \(0\le n\le N\) (\(N\) at least \(7\)) and \(r_1\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(N>7\).

C \(x^2(1+2x)y''+x(5+9x)y'+(4+3x)y=0\)

Show answer

\(\dst{y_1=x^{-2}\left(1+3x+{3\over2}x^2-{1\over2}x^3+\cdots\right)}\); \(\dst{y_2=y_1\ln x-5x^{-1}\left(1+{5\over4}x-{1\over4}x^2+\cdots\right)}\)
C \(x^2(1-2x)y''-x(5+4x)y'+(9+4x)y=0\)

Show answer

\(\dst{y_1=x^3(1+20x+180x^2+1120x^3+\cdots}\); \(\dst{y_2=y_1\ln x-x^4\left(26+324x+{6968\over3}x^2+\cdots\right)}\)
C \(x^2(1+4x)y''-x(1-4x)y'+(1+x)y=0\)

Show answer

\(\dst{y_1=x\left(1-5x+{85\over4}x^2-{3145\over36}x^3+\cdots\right)}\); \(\dst{y_2= y_1\ln x+x^2\left(2-{39\over4}x+{4499\over108}x^2+\cdots\right)}\)
C \(x^2(1+x)y''+x(1+2x)y'+xy=0\)

Show answer

\(\dst{y_1=1-x+{3\over4}x^2-{7\over12}x^3+\cdots}\); \(\dst{y_2=y_1\ln x+x\left(1-{3\over4}x+{5\over9}x^2+\cdots\right)}\)
C \(x^2(1-x)y''+x(7+x)y'+(9-x)y=0\)

Show answer

\(\dst{y_1=x^{-3}(1+16x+36x^2+16x^3+\cdots)}\); \(\dst{y_2=y_1\ln x-x^{-2}\left(40+150x+{280\over3}x^2+\cdots\right)}\)

In Exercises 28–38 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

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

Show answer

\(\dst{y_1=x\sum_{m=0}^\infty{(-1)^m\over2^mm!}x^{2m}}\); \(\dst{y_2=y_1\ln x-{x\over2}\sum_{m=1}^\infty{(-1)^m\over 2^mm!}\left(\sum_{j=1}^m{1\over j}\right)x^{2m}}\)
\(x^2(1+x^2)y''-3x(1-x^2)y'+4y=0\)

Show answer

\(\dst{y_1=x^2\sum_{m=0}^\infty(-1)^m(m+1)x^{2m}}\); \(\dst{y_2=y_1\ln x-{x^2\over2}\sum_{m=1}^\infty(-1)^mmx^{2m}}\)
\(4x^2y''+2x^3y'+(1+3x^2)y=0\)

Show answer

\(\dst{y_1=x^{1/2}\sum_{m=0}^\infty{(-1)^m\over4^mm!}x^{2m}}\); \(\dst{y_2=y_1\ln x-{x^{1/2}\over2}\sum_{m=1}^\infty{(-1)^m\over 4^mm!}\left(\sum_{j=1}^m{1\over j}\right) x^{2m}}\)
\(x^2(1+x^2)y''-x(1-2x^2)y'+y=0\)

Show answer

\(\dst{y_1=x\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(2j-1)\over2^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x+{x\over2}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(2j-1)\over2^mm!} \left(\sum_{j=1}^m{1\over j(2j-1)}\right)x^{2m}}\)
\(2x^2(2+x^2)y''+7x^3y'+(1+3x^2)y=0\)

Show answer

\(\dst{y_1=x^{1/2}\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(4j-1)\over8^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x+{x^{1/2}\over2}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(4j-1)\over8^mm!} \left(\sum_{j=1}^m{1\over j(4j-1)}\right)x^{2m}}\)
\(x^2(1+x^2)y''-x(1-4x^2)y'+(1+2x^2)y=0\)

Show answer

\(\dst{y_1=x\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(2j+1)\over2^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x-{x\over2}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(2j+1)\over2^mm!} \left(\sum_{j=1}^m{1\over j(2j+ 1)}\right)x^{2m}}\)
\(4x^2(4+x^2)y''+3x(8+3x^2)y'+(1-9x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/4}\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(8j-13)\over(32)^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x+{13\over2}x^{-1/4}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(8j-13)\over(32)^mm!} \left(\sum_{j=1}^m{1\over j(8j-13)}\right)x^{2m}}\)
\(3x^2(3+x^2)y''+x(3+11x^2)y'+(1+5x^2)y=0\)

Show answer

\(\dst{y_1=x^{1/3}\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(3j-1)\over9^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x+{x^{1/3}\over2}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(3j-1)\over9^mm!} \left(\sum_{j=1}^m{1\over j(3j-1)}\right)x^{2m}}\)
\(4x^2(1+4x^2)y''+32x^3y'+y=0\)

Show answer

\(\dst{y_1=x^{1/2}\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(4j-3)(4j-1)\over4^m(m!)^2} x^{2m}}\);

\(\dst{y_2=y_1\ln x+x^{1/2}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(4j-3)(4j-1)\over4^m(m!)^2} \left(\sum_{j=1}^m{8j-3\over j(4j-3) (4j-1)}\right)x^{2m}}\)
\(9x^2y''-3x(7-2x^2)y'+(25+2x^2)y=0\)

Show answer

\(\dst{y_1=x^{5/3}\sum_{m=0}^\infty{(-1)^m\over3^mm!}x^{2m}}\); \(\dst{y_2=y_21\ln x-{x^{5/3}\over2}\sum_{m=1}^\infty{(-1)^m\over3^mm!}\left(\sum_{j=1}^m{1\over j}\right)x^{2m}}\)
\(x^2(1+2x^2)y''+x(3+7x^2)y'+(1-3x^2)y=0\)

Show answer

\(\dst{y_1={1\over x }\sum_{m=0}^\infty{(-1)^m\prod_{j=1}^m(4j-7)\over2^mm!} x^{2m}}\);

\(\dst{y_2=y_1\ln x+{7\over2x}\sum_{m=1}^\infty{(-1)^m\prod_{j=1}^m(4j-7)\over2^mm!} \left(\sum_{j=1}^m{1\over j(4j-7)}\right)x^{2m}}\)

In Exercises 39–43 find a fundamental set of Frobenius solutions. Compute the terms involving \(x^{2m+r_1}\), where \(0\le m\le M\) (\(M\) at least \(3\)) and \(r_1\) is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take \(M>3\).

C \(x^2(1+x^2)y''+x(3+8x^2)y'+(1+12x^2)y\)

Show answer

\(\dst{y_1=x^{-1}\left(1-{3\over2}x^2+{15\over8}x^4-{35\over16}x^6+\cdots\right)}\)

; \(\dst{y_2=y_1\ln x+x\left({1\over4}-{13\over32}x^2+{101\over192}x^4 +\cdots\right)}\)
C \(x^2y''-x(1-x^2)y'+(1+x^2)y=0\)

Show answer

\(\dst{y_1=x\left(1-{1\over2}x^2+{1\over8}x^4-{1\over48}x^6+\cdots\right)}\); \(\dst{y_2=y_1 \ln x+x^3\left({1\over4}-{3\over32}x^2+ {11\over576}x^4+\cdots\right)}\)
C \(x^2(1-2x^2)y''+x(5-9x^2)y'+(4-3x^2)y=0\)

Show answer

\(\dst{y_1=x^{-2}\left(1-{3\over4}x^2-{9\over64}x^4-{25\over256}x^6 +\cdots\right)}\); \(\dst{y_2=y_1\ln x+{1\over2}-{21\over128}x^2-{215\over1536}x^4+\cdots}\)
C \(x^2(2+x^2)y''+x(14-x^2)y'+2(9+x^2)y=0\)

Show answer

\(\dst{y_1=x^{-3}\left(1-{17\over8}x^2+{85\over256}x^4-{85\over18432}x^6+\cdots\right)}\); \(\dst{y_2=y_1\ln x +x^{-1}\left({25\over8}-{471\over512}x^2+{1583\over110592}x^4 +\cdots\right)}\)
C \(x^2(1+x^2)y''+x(3+7x^2)y'+(1+8x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1}\left(1-{3\over4}x^2+{45\over64}x^4-{175\over256}x^6+\cdots \right)}\); \(\dst{y_2=y_1\ln x-x\left({1\over4}-{33\over128}x^2+{395\over1536}x^4 +\cdots\right)}\)

In Exercises 44–52 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

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

Show answer

\(\dst{y_1={1\over x}}\); \(\dst{y_2=y_1\ln x-6+6x-{8\over3}x^2}\)
\(x(1+x)y''+(1-x)y'+y=0\)

Show answer

\(\dst{y_1=1-x}\); \(\dst{y_2=y_1\ln x+4x}\)
\(x^2(1-x)y''+x(3-2x)y'+(1+2x)y=0\)

Show answer

\(\dst{y_1={(x-1)^2\over x}}\); \(\dst{y_2=y_1\ln x+3-3x+2\sum_{n=2}^\infty {1\over n(n^2-1)}x^n}\)
\(4x^2(1+x)y''-4x^2y'+(1-5x)y=0\)

Show answer

\(\dst{y_1=x^{1/2}(x+1)^2}\); \(\dst{y_2=y_1\ln x-x^{3/2}\left( 3+3x+2\sum_{n=2}^\infty{(-1)^n\over n(n^2-1)}x^n\right)}\)
\(x^2(1-x)y''-x(3-5x)y'+(4-5x)y=0\)

Show answer

\(\dst{y_1=x^2(1-x)^3}\); \(\dst{y_2=y_1\ln x+x^3\left(4-7x+{11\over3}x^2-6\sum_{n=3}^\infty{1\over n(n-2)(n^2-1)}x^n\right)}\)
\(x^2(1+x^2)y''-x(1+9x^2)y'+(1+25x^2)y=0\)

Show answer

\(\dst{y_1=x-4x^3+x^5}\); \(\dst{y_2=y_1\ln x+6x^3-3x^5}\)
\(9x^2y''+3x(1-x^2)y'+(1+7x^2)y=0\)

Show answer

\(\dst{y_1=x^{1/3}\left(1-{1\over6}x^2\right)}\); \(\dst{y_2=y_1\ln x +x^{7/3}\left({1\over4}-{1\over12}\sum_{m=1}^\infty {1\over6^mm(m+1)(m+1)!}x^{2m}\right)}\)
\(x(1+x^2)y''+(1-x^2)y'-8xy=0\)

Show answer

\(\dst{y_1=(1+x^2)^2}\); \(\dst{y_2=y_1\ln x-{3\over2}x^2-{3\over2}x^4 +\sum_{m=3}^\infty{(-1)^m\over m(m-1)(m-2)}x^{2m}}\)
\(4x^2y''+2x(4-x^2)y'+(1+7x^2)y=0\)

Show answer

\(\dst{y_1=x^{-1/2}\left(1-{1\over2}x^2+{1\over32}x^4\right)}\); \(\dst{y_2=y_1\ln x+x^{3/2}\left({5\over8}-{9\over128}x^2 +\sum_{m=2}^\infty{1\over4^{m+1}(m-1)m(m+1)(m+1)!}x^{2m}\right)}\).
Under the assumptions of Theorem 7.6.2, suppose the power series

\[ \sum_{n=0}^\infty a_n(r_1)x^n \quad\mbox{ and }\quad \sum_{n=1}^\infty a_n'(r_1)x^n \]

converge on \((-\rho,\rho)\).
1. Show that
  
  \[ y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n \]
  
  are linearly independent on \((0,\rho)\).
  
  Hint
  
  Show that if \(c_1\) and \(c_2\) are constants such that \(c_1y_1+c_2y_2\equiv0\) on \((0,\rho)\), then
  
  \[ (c_1+c_2\ln x)\sum_{n=0}^\infty a_n(r_1)x^n+ c_2\sum_{n=1}^\infty a_n'(r_1)x^n=0,\quad 0<x<\rho. \]
  
  Then let \(x\to0+\) to conclude that \(c_2=0\).
2. Use the result of (a) to complete the proof of Theorem 7.6.2.
Let

\[ Ly=x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y \]

and define

\[ p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_1(r)=\alpha_1r(r-1)+\beta_1r+\gamma_1. \]

Theorem 7.6.1 and Exercise 7.5. 55(a) imply that if

\[ y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n \]

where

\[ a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)}, \]

then

\[ Ly(x,r)=p_0(r)x^r. \]

Now suppose \(p_0(r)=\alpha_0(r-r_1)^2\) and \(p_1(k+r_1)\ne0\) if \(k\) is a nonnegative integer.
1. Show that \(Ly=0\) has the solution
  
  \[ y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n, \]
  
  where
  
  \[ a_n(r_1)={(-1)^n\over\alpha_0^n(n!)^2}\prod_{j=1}^np_1(j+r_1-1). \]
2. Show that \(Ly=0\) has the second solution
  
  \[ y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n(r_1)J_nx^n, \]
  
  where
  
  \[ J_n=\sum_{j=1}^n{p_1'(j+r_1-1)\over p_1(j+r_1-1)}-2\sum_{j=1}^n{1\over j}. \]
3. Conclude from (a) and (b) that if \(\gamma_1\ne0\) then
  
  \[ y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^nx^n \]
  
  and
  
  \[ y_2=y_1\ln x-2x^{r_1}\sum_{n=1}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{1\over j}\right)x^n \]
  
  are solutions of
  
  \[ \alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0. \]
  
  (The conclusion is also valid if \(\gamma_1=0\). Why?)
Let

\[ Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+(\gamma_0+\gamma_qx^q)y \]

where \(q\) is a positive integer, and define

\[ p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_q(r)=\alpha_qr(r-1)+\beta_qr+\gamma_q. \]

Suppose

\[ p_0(r)=\alpha_0(r-r_1)^2 \quad\mbox{ and }\quad p_q(r)\not\equiv0. \]
1. Recall from Exercise 7.5. 59 that \(Ly~=0\) has the solution
  
  \[ y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm}, \]
  
  where
  
  \[ a_{qm}(r_1)={(-1)^m\over (q^2\alpha_0)^m(m!)^2}\prod_{j=1}^mp_q\left(q(j-1)+r_1\right). \]
2. Show that \(Ly=0\) has the second solution
  
  \[ y_2=y_1\ln x+x^{r_1}\sum_{m=1}^\infty a_{qm}'(r_1)J_mx^{qm}, \]
  
  where
  
  \[ J_m=\sum_{j=1}^m{p_q'\left(q(j-1)+r_1\right)\over p_q\left(q(j-1)+r_1\right)}-{2\over q}\sum_{j=1}^m{1\over j}. \]
3. Conclude from (a) and (b) that if \(\gamma_q\ne0\) then
  
  \[ y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^mx^{qm} \]
  
  and
  
  \[ y_2=y_1\ln x-{2\over q}x^{r_1}\sum_{m=1}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^m\left(\sum_{j=1}^m{1\over j}\right)x^{qm} \]
  
  are solutions of
  
  \[ \alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0. \]
The equation

\[ xy''+y'+xy=0 \]

is Bessel’s equation of order zero. (See Exercise 53.) Find two linearly independent Frobenius solutions of this equation.

Show answer

\(\dst{y_1=\sum_{m=0}^\infty{(-1)^m\over4^m(m!)^2}x^{2m}}\); \(\dst{y_2=y_1\ln x-\sum_{m=1}^\infty{(-1)^m\over4^m(m!)^2}\left(\sum_{j=1}^m{1\over j}\right)x^{2m}}\)
Suppose the assumptions of Exercise 7.5. 53 hold, except that

\[ p_0(r)=\alpha_0(r-r_1)^2. \]

Show that

\[ y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_1}\ln x\over\alpha_0+\alpha_1x+\alpha_2x^2} \]

are linearly independent Frobenius solutions of

\[ x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0 \]

on any interval \((0,\rho)\) on which \(\alpha_0+\alpha_1x+\alpha_2x^2\) has no zeros.

In Exercises 58–65 use the method suggested by Exercise 57 to find the general solution on some interval \((0,\rho)\).

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

Show answer

\(\dst{x^{1/2}\over1+x}\); \(\dst{x^{1/2}\ln x\over1+x}\)
\(9x^2(3+x)y''+3x(3+7x)y'+(3+4x)y=0\)

Show answer

\(\dst{x^{1/3}\over3+x}\); \(\dst{x^{1/3}\ln x\over3+x}\)
\(x^2(2-x^2)y''-x(2+3x^2)y'+(2-x^2)y=0\)

Show answer

\(\dst{x\over2-x^2}\); \(\dst{x\ln x\over2-x^2}\)
\(16x^2(1+x^2)y''+8x(1+9x^2)y'+(1+49x^2)y=0\)

Show answer

\(\dst{x^{1/4}\over1+x^2}\); \(\dst{x^{1/4}\ln x\over1+x^2}\)
\(x^2(4+3x)y''-x(4-3x)y'+4y=0\)

Show answer

\(\dst{x\over4+3x}\); \(\dst{x\ln x\over4+3x}\)
\(4x^2(1+3x+x^2)y''+8x^2(3+2x)y'+(1+3x+9x^2)y=0\)

Show answer

\(\dst{x^{1/2}\over1+3x+x^2}\); \(\dst{x^{1/2}\ln x\over1+3x+x^2}\)
\(x^2(1-x)^2y''-x(1+2x-3x^2)y'+(1+x^2)y=0\)

Show answer

\(\dst{x\over(1-x)^2}\); \(\dst{x\ln x\over(1-x)^2}\)
\(9x^2(1+x+x^2)y''+3x(1+7x+13x^2)y'+(1+4x+25x^2)y=0\)

Show answer

\(\dst{x^{1/3}\over1+x+x^2}\); \(\dst{x^{1/3}\ln x\over1+x+x^2}\)
1. Let \(L\) and \(y(x,r)\) be as in Exercises 57 and 58. Extend Theorem 7.6.1 by showing that
  
  \[ L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x. \]
2. Show that if
  
  \[ p_0(r)=\alpha_0(r-r_1)^2 \]
  
  then
  
  \[ y_1=y(x,r_1) \mbox{\quad and \quad} y_2={\partial y\over\partial r}(x,r_1) \]
  
  are solutions of \(Ly=0\).