M555: Differential Equations

5.4 Regular Singular Points, Part II

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5.4.5 Series Solutions Near a Regular Singular Point

We seek a solution of the differential equation

$$x^2y'' + x\left(xp(x)\right)y' + \left(x^2q(x)\right)y = 0$$
with variable coefficients that satisfy the conditions

$$xp(x) =\color{cornflowerblue}{\displaystyle\sum_{n=0}^{\infty} p_nx^n }\qquad\qquad\qquad x^2 q(x) =\color{magenta}{\displaystyle\sum_{n=0}^{\infty} q_nx^n } $$
of the form

$$y(x) = \phi(r,x) = \color{orange}{\displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}}.$$
Substituting these expressions into the differential equation we obtain

$$\begin{align*} x^2\color{teal}{\displaystyle\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2}} + x &\left(\color{cornflowerblue}{\displaystyle\sum_{n=0}^{\infty} p_nx^n}\right)\left(\color{red}{\displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1}}\right) +\ldots \\ &\left(\color{magenta}{\displaystyle\sum_{n=0}^{\infty} q_nx^n}\right)\left(\color{orange}{\displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}}\right) = 0 \end{align*}$$
Distributing the variable $x$ across the absolutely convergent series results in

$$\begin{align*} \color{teal}{\displaystyle\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r}} + &\left(\color{cornflowerblue}{\displaystyle\sum_{n=0}^{\infty} p_nx^n}\right)\left(\color{red}{\displaystyle\sum_{n=0}^{\infty} a_n(n+r)x^{n+r}}\right) +\ldots \\ &\left(\color{magenta}{\displaystyle\sum_{n=0}^{\infty} q_nx^n}\right)\left(\color{orange}{\displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}}\right) = 0 \end{align*}$$
If we define our linear operator $L[y] = x^2y'' + x(xp(x))y' + x^2q(x)y$, then we can write out the differential equation using ellipses

$$\begin{align*} L[y] &= L[\phi(r,x)] \\ &= \left(\color{teal}{a_0r(r-1)x^r + a_1(r+1)rx^{r+1}+\cdots+a_n(n+r)(n+r-1)x^{n+r-2}+\cdots}\right) \\ &\quad + \left(\color{cornflowerblue}{p_0+p_1x+\cdots+p_nx^n+\cdots}\right)\left(\color{red}{a_0rx^r + a_1(r+1)x^{r+1}+\cdots+a_n(n+r)x^{n+r}+\cdots}\right) \\ &\quad + \left(\color{magenta}{q_0+q_1x+\cdots+q_nx^n+\cdots}\right)\left(\color{orange}{a_0x^r+a_1x^{r+1}+\cdots+a_nx^{n+r}+\cdots}\right) \end{align*}$$
If we expand the multiplication and group together terms with the same power of $x$,

$$\begin{align*} L[\phi] &= \left[\color{teal}{a_0r(r-1)x^r} + \color{cornflowerblue}{p_0}\color{red}{a_0rx^r} + \color{magenta}{q_0}\color{orange}{a_0x^r}\right] \\ &\qquad + \left[\color{teal}{a_1(r+1)rx^{r+1}} + \color{cornflowerblue}{p_0}\color{red}{a_1(r+1)x^{r+1}} + \color{cornflowerblue}{p_1x}\color{red}{a_0rx^r} + \color{magenta}{q_0}\color{orange}{a_1x^{r+1}} + \color{magenta}{q_1x}\color{orange}{a_0x^r}\right] \\ &\qquad + \left[\color{teal}{a_2(r+2)(r+1)x^{r+2}} + \color{cornflowerblue}{p_0}\color{red}{a_2(r+2)x^{r+2}} + \color{cornflowerblue}{p_1x}\color{red}{a_1(r+1)x^{r+1}} + \color{cornflowerblue}{p_2x^2}\color{red}{a_0rx^r}\right. \\ &\qquad\qquad + \left.\color{magenta}{q_0}\color{orange}{a_2x^{r+2}} + \color{magenta}{q_1x}\color{orange}{a_1x^{r+1}} + \color{magenta}{q_2x^2}\color{orange}{a_0x^r}\right] \\ &\qquad\ddots \\ &\qquad + \left[\color{teal}{a_n(r+n)(r+n-1)x^{r+n}} + \color{cornflowerblue}{p_0}\color{red}{a_n(r+n)x^{r+n}} + \color{cornflowerblue}{p_1x}\color{red}{a_{n-1}(r+n-1)x^{r+n-1}} +\cdots \right. \\ &\qquad\qquad \left.+\ \color{cornflowerblue}{p_nx^n}\color{red}{a_0rx^r} + \color{magenta}{q_0}\color{orange}{a_nx^{r+n}} + \color{magenta}{q_1x}\color{orange}{a_{n-1}x^{r+n-1}} +\cdots+ \color{magenta}{q_nx^n}\color{orange}{a_0x^r} \right] \\ &\qquad\ddots \\ \end{align*}$$
Now we distribute the $a_n$'s and powers of $x$ to simplify each line,

$$\begin{align*} L[\phi] &= a_0\left[\color{teal}{r(r-1)} + \color{cornflowerblue}{p_0}\color{red}{r} + \color{magenta}{q_0}\right]x^r \\ &\qquad + \left[a_1\left[\color{teal}{(r+1)r} + \color{cornflowerblue}{p_0}\color{red}{(r+1)} + \color{magenta}{q_0}\right] + a_0\left(\color{cornflowerblue}{p_1}\color{red}{r} + \color{magenta}{q_1}\right)\right]x^{r+1} \\ &\qquad + \left[a_2\left[\color{teal}{(r+2)(r+1)} + \color{cornflowerblue}{p_0}\color{red}{(r+2)} + \color{magenta}{q_0}\right] + a_1\left(\color{cornflowerblue}{p_1}\color{red}{(r+1)} + \color{magenta}{q_1}\right) + a_0\left(\color{cornflowerblue}{p_2}\color{red}{r} + \color{magenta}{q_2}\right)\right]x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left[a_n\left[\color{teal}{(r+n)(r+n-1)} + \color{cornflowerblue}{p_0}\color{red}{(r+n)} + \color{magenta}{q_0}\right] + a_{n-1}\left(\color{cornflowerblue}{p_1}\color{red}{(r+n-1)} + \color{magenta}{q_1}\right) +\cdots \right. \\ &\qquad\qquad \left.+\ a_0\left(\color{cornflowerblue}{p_n}\color{red}{r} + \color{magenta}{q_n}\right) \right]x^{r+n} \\ &\qquad\ddots \\ \end{align*}$$
If we define

$$F(r) = \left[\color{teal}{r(r-1)} + \color{cornflowerblue}{p_0}\color{red}{r} + \color{magenta}{q_0}\right],$$
then we have

$$\begin{align*} L[\phi] &= a_0F(r)x^r + \left[a_1F(r+1) + a_0\left(\color{cornflowerblue}{p_1}\color{red}{r} + \color{magenta}{q_1}\right)\right]x^{r+1} \\ &\qquad + \left[a_2F(r+2) + a_1\left(\color{cornflowerblue}{p_1}\color{red}{(r+1)} + \color{magenta}{q_1}\right) + a_0\left(\color{cornflowerblue}{p_2}\color{red}{r} + \color{magenta}{q_2}\right)\right]x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left[a_nF(r+n) + a_{n-1}\left(\color{cornflowerblue}{p_1}\color{red}{(r+n-1)} + \color{magenta}{q_1}\right) +\cdots + a_0\left(\color{cornflowerblue}{p_n}\color{red}{r} + \color{magenta}{q_n}\right) \right]x^{r+n} \\ &\qquad\ddots \\ \end{align*}$$
Now we arrange the terms after $a_nF(r+n)$ to increase from $0$ to $n-1$

$$\begin{align*} L[\phi] &= a_0F(r)x^r + \left[a_1F(r+1) + a_0\left(\color{cornflowerblue}{p_1}\color{red}{r} + \color{magenta}{q_1}\right)\right]x^{r+1} \\ &\qquad + \left[a_2F(r+2) + a_0\left(\color{cornflowerblue}{p_2}\color{red}{r} + \color{magenta}{q_2}\right) + a_1\left(\color{cornflowerblue}{p_1}\color{red}{(r+1)} + \color{magenta}{q_1}\right)\right]x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left[a_nF(r+n) + a_0\left(\color{cornflowerblue}{p_n}\color{red}{r} + \color{magenta}{q_n}\right) +\cdots + a_{n-1}\left(\color{cornflowerblue}{p_1}\color{red}{(r+n-1)} + \color{magenta}{q_1}\right)\right]x^{r+n} \\ &\qquad\ddots \\ \end{align*}$$
If we write the differential equation with sum signs instead of ellipses

$$L[\phi] = a_0F(r)x^r + \displaystyle\sum_{n=1}^{\infty}\left( a_nF(r+n) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right) \right)x^{r+n} = 0.$$
Notice that since the coefficient of $x^r$ must be zero and $a_0\neq 0$, we have the indicial equation

$$F(r) = r(r-1) + p_0r + q_0 = 0.$$
Since all of the coefficients must be zero as well, we have the recurrence equation

$$a_nF(r+n) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right) = 0,\qquad n > 0.$$
Thus the coefficient $a_n$ depends on $r$ and all of the previous coefficients. This gives us the rather frightening version of the recurrence relation

$$a_n(r) = -\dfrac{\displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right)}{F(r+n)}.$$

Theorem 5.6.1

For the differential equation

$$x^2y'' + x\left(xp(x)\right)y' + \left(x^2q(x)\right)y = 0,$$
with a regular singular at $x=0$, and where $xp(x)$ and $x^2q(x)$ are analytic on some interval of radius $\rho$ centered at $0$. Then each has an absolutely convergent power series,

$$xp(x) = \displaystyle\sum_{n=0}^{\infty} p_nx^n$$
and

$$x^2q(x) = \displaystyle\sum_{n=0}^{\infty} q_nx^n.$$

Let $r_1\ge r_2$ be the real roots of the indicial equation

$$F(r) = r(r-1) + p_0r + q_0.$$
Then on either interval $(-\rho,0)$ or $(0,\rho)$, the first fundamental solution is of the form
$$y_1(x) = |x|^{r_1}\left(1 + \displaystyle\sum_{n=1}^{\infty} a_n(r_1)x^n\right),$$
where $a_0=1$ and the remaining coefficients are given by the recurrence equation with $r=r_1$

$$a_n(r_1)F(r_1+n) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(r_1+k) + q_{n-k}\right) = 0,\qquad n > 0.$$ This power series for $y_1$ will converge at least for $|x|<\rho$, but not necessarily at $x=0$.

Case 1:

If $r_1>r_2$ are distinct and $r_1-r_2$ is not a positive integer, then the second fundamental solution is given by

$$y_2(x) = |x|^{r_2}\left(1 + \displaystyle\sum_{n=1}^{\infty} a_n(r_2)x^n\right),$$
where $a_0=1$ and the remaining coefficients are given by the recurrence equation with $r=r_2$

$$a_n(r_2)F(r_2+n) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(r_2+k) + q_{n-k}\right) = 0,\qquad n > 0.$$

Case 2:

If $r_1=r_2$, then the second fundamental solution is given by

$$y_2(x) = y_1(x)\ln|x| + |x|^{r_1}\left(1 + \displaystyle\sum_{n=1}^{\infty} b_n(r_1)x^n\right).$$

Case 3:

If $r_1-r_2$ is a positive integer, then

$$y_2(x) = ay_1(x)\ln|x| + |x|^{r_2}\left(1 + \displaystyle\sum_{n=1}^{\infty} c_n(r_2)x^n\right).$$
The coefficients $a_n$, $b_n$ and $c_n$, and the constant $a$ can be determined by substituting the equation for $y_2$ for $y$ in the differential equation , or use reduction of order as we have done in previous examples in this section. Each of these series for $y_2$ converges absolutely for at least $|x|<\rho$, but not necessarily at $x=0$.

Problem 5.4.5

For the differential equation

$$xy'' + 2xy' + 6e^xy = 0,\qquad x \gt 0.$$

1. Show that $x=0$ is a regular singular point,
   
   View Solution

   
   $P(x)=x$, so $P(0)=0$ is the only singular point. The standard form of our equation is
   $$x^2y'' + x(2x)y' + x\left(6e^x\right)y = 0,\qquad x \gt 0.$$
   So we have that
   $$p(x) = 2x\ \text{, and }\ q(x) = \dfrac{6e^x}{x}.$$
   $$\displaystyle\lim_{x\to 0} xp(x) = \displaystyle\lim_{x\to 0} 2x^2 = 0,$$
   $$\displaystyle\lim_{x\to 0} x^2q(x) = \displaystyle\lim_{x\to 0} x^2\dfrac{6e^x}{x} = \displaystyle\lim_{x\to 0} 6xe^x = 0.$$
   Therefore $x=0$ is a regular singular point.
   
   
2. Find the exponents at the singular point $x=0$.
   
   View Solution

   The power series for $xp(x)$ and $x^2q(x)$ are $$\begin{align*} xp(x) &= 0 + 2x + 0 + 0 +\cdots+ 0 +\cdots,\\ x^2q(x) &= 6x\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = \sum_{n=0}^{\infty}\dfrac{6}{n!}x^{n+1}\\ &= 0 + 6x + 6x^2 + 3x^3 + x^4 +\cdots+ \dfrac{6}{n!}x^{n+1} +\cdots. \end{align*} $$
   Thus $p_0 = q_0 = 0$. The indical equation becomes
   $$F(r) = r(r-1)=0.$$
   So the exponents at the singular point $x=0$ are $r_1=1$ and $r_2=0$.
   
   
3. Find the first four non-zero terms of the series of the first fundamental solutions about $x=0$.
   
   View Solution

   The first fundamental solution is found using the recurrence equation and $r=1$,
   $$ a_nF(n+1) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(k+1) + q_{n-k}\right) = 0,\qquad n \gt 0,$$
   or
   $$a_n = -\dfrac{\displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}(k+1) + q_{n-k}\right)}{n(n+1)},\qquad n \gt 0.$$
   Thus we have
   $$\begin{align*} a_0 &= 1, \\ \\ a_1 &= -\dfrac{a_0\left(p_1(1) + q_1\right)}{1(2)} = -\dfrac{1(2 + 6)}{2} = -4, \\ \\ a_2 &= -\dfrac{a_0\left(p_2(1)+q_2\right) + a_1\left(p_1(2) + q_1\right)}{2(3)} = -\dfrac{1(0+6) - 4(2(2)+6)}{6} = -\dfrac{6-40}{6} = \dfrac{17}{3}, \\ \\ a_3 &= -\dfrac{a_0\left(p_3(1)+q_3\right) + a_1\left(p_2(2) + q_2\right) + a_2\left(p_1(3) + q_1\right)}{3(4)} \\ \\ &= -\dfrac{1(0+3) - 4(0(2) + 6) + \frac{17}{3}(2(3) + 6)}{12} = -\dfrac{3 - 24 + 68}{12} = -\dfrac{47}{12}. \end{align*}$$
   Substituting these values into the equation for the first fundamental solution results in
   
   $$\begin{align*} y_1(x) &= x^1\left(1 - 4x + \dfrac{17}{3}x^2 - \dfrac{47}{12}x^3 +\cdots \right) \\ &= x + 4x^2 + \dfrac{17}{3}x^3 - \dfrac{47}{12}x^4 +\cdots. \end{align*}$$
   
   
   

Problem 5.4.6

For the differential equation

$$xy'' + y' - y = 0,\qquad x>0.$$

1. Show that $x=0$ is a regular singular point,
   
   
   

   View Solution

   $P(x)=x$, so $P(0)=0$ is the only singular point. The standard form of our equation is
   $$x^2y'' + x(1)y' + x\left(-1\right)y = 0,\qquad x>0.$$
   So we have that
   $$p(x) = \dfrac{1}{x}\ \text{, and }\ q(x) = -\dfrac{1}{x}.$$
   $$\begin{align*} \displaystyle\lim_{x\to 0} xp(x) &= \displaystyle\lim_{x\to 0} x\left(\dfrac{1}{x}\right) = \displaystyle\lim_{x\to 0} 1 = 1, \\ \\ \displaystyle\lim_{x\to 0} x^2q(x) &= \displaystyle\lim_{x\to 0} x^2\left(-\dfrac{1}{x}\right) = \displaystyle\lim_{x\to 0} -x = 0. \\ \end{align*}$$
   Therefore $x=0$ is a regular singular point.
   
   

2. Find the exponents at the singular point $x=0$.
   

   
   

   View Solution

   The power series for $xp(x)$ and $x^2q(x)$ are
   
   $$\begin{align*} xp(x) &= 1 + 0 + 0 + 0 +\cdots, \\ \\ x^2q(x) &= 0 - x + 0 + 0 +\cdots \end{align*}$$
   Thus $p_0 = 1$ and $q_0 = 0$. The indicial equation becomes
   $$F(r) = r(r-1) + r = r^2 = 0.$$
   So the exponents at the singular point $x=0$ are $r_1=r_2=0$.
   
   

3. Find the first four non-zero terms of the series of the first fundamental solutions about $x=0$.
   

   
   

   View Solution

   The first fundamental solution is given by the recurrence equation and $r=0$,
   $$a_nF(n) + \displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}k + q_{n-k}\right) = 0,\qquad n \gt 0,$$
   or
   $$a_n = -\dfrac{\displaystyle\sum_{k=0}^{n-1} a_k\left(p_{n-k}k + q_{n-k}\right)}{n^2},\qquad n \gt 0.$$
   Thus we have
   $$\begin{align*} a_0 &= 1, \\ \\ a_1 &= -\dfrac{a_0\left(p_1(0) + q_1\right)}{1^2} = -\left(2(0-1)\right) = 1, \\ \\ a_2 &= -\dfrac{a_0\left(p_2(0)+q_2\right) + a_1\left(p_1(1) + q_1\right)}{2^2} = -\dfrac{1(0+0) + 1(0(1)-1)}{4} = \dfrac{1}{4}, \\ \\ a_3 &= -\dfrac{a_0\left(p_3(0)+q_3\right) + a_1\left(p_2(1) + q_2\right) + a_2\left(p_1(2) + q_1\right)}{3^2} \\ &= -\dfrac{1(0+0) + 1(0(1) + 0) + \frac{1}{4}(0(2) - 1)}{9} = -\dfrac{0 + 0 - \frac{1}{4}}{9} = \dfrac{1}{36},\\ \\ a_4 &= -\dfrac{a_0\left(p_4(0)+q_4\right) + a_1\left(p_3(1) + q_3\right) + a_2\left(p_2(2) + q_2\right) + a_3\left(p_1(3) + q_1\right)}{4^2} \\ &= -\dfrac{1(0+0) + 1(0(1)+0) + \frac{1}{4}(0(2)+0) + \frac{1}{36}(0(3)-1)}{16} = \dfrac{1}{576}. \end{align*}$$
   Substituting these values into the equation for the first fundamental solution results in
   
   $$\begin{align*} y_1(x) &= x^0\left(1 + x + \dfrac{1}{4}x^2 + \dfrac{1}{36}x^3 + \dfrac{1}{576}x^4 +\cdots \right)\\ \\ &= 1 + x + \dfrac{1}{4}x^2 + \dfrac{1}{36}x^3 + \dfrac{1}{576}x^4 +\cdots. \end{align*}$$
   From theorem 3.6.1 we have case 2 and the second fundamental solution has the form
   
   $$y_2(x) = y_1(x)\ln|x| + |x|^{r_1}\displaystyle\sum_{n=1}^{\infty} b_n(r_1)x^n.$$
   Substituting $r_1=0$ and differentiating we obtain for $x>0$
   $$\begin{align*} y_2(x) &= y_1\ln(x) + \displaystyle\sum_{n=1}^{\infty} b_nx^n \\ \\ y_2'(x) &= y_1'\ln(x) + \dfrac{1}{x}y_1 + \displaystyle\sum_{n=1}^{\infty} nb_nx^{n-1} \\ \\ y_2''(x) &= y_1''\ln(x) + \dfrac{2}{x}y_1' - \dfrac{1}{x^2}y_1 + \displaystyle\sum_{n=2}^{\infty} n(n-1)b_nx^{n-2} \\ \\ xy_2''(x) &= y_1''x\ln(x) + 2y_1' - \dfrac{1}{x}y_1 + \displaystyle\sum_{n=2}^{\infty} n(n-1)b_nx^{n-1} \\ \\ \end{align*}$$
   $$\begin{align*} xy'' + y' - y &= y_1''x\ln(x) + 2y_1' - \dfrac{1}{x}y_1 + \displaystyle\sum_{n=2}^{\infty} n(n-1)b_nx^{n-1} \\ &\qquad + y_1'\ln(x) + \dfrac{1}{x}y_1 + \displaystyle\sum_{n=1}^{\infty} nb_nx^{n-1} - \left(y_1\ln(x) + 1 + \displaystyle\sum_{n=1}^{\infty} b_nx^n\right) \\ \\ &= \left(xy_1'' + y_1' - y_1\right)\ln(x) + 2y_1' + \left(-\dfrac{1}{x} + \dfrac{1}{x}\right)y_1 \\ &\qquad + \displaystyle\sum_{n=2}^{\infty} n(n-1)b_nx^{n-1} + \displaystyle\sum_{n=1}^{\infty} nb_nx^{n-1} - \displaystyle\sum_{n=1}^{\infty} b_nx^n \\ \\ &= (0)\ln(x) + 2y_1' + (0)y_1 + \displaystyle\sum_{n=1}^{\infty} (n+1)nb_{n+1}x^n + \sum_{n=0}^{\infty} (n+1)b_{n+1}x^n - \displaystyle\sum_{n=1}^{\infty} b_nx^n \\ \\ &= 2y_1' + (1)b_1 + \displaystyle\sum_{n=1}^{\infty} (n+1)nb_{n+1}x^n + \sum_{n=1}^{\infty} (n+1)b_{n+1}x^n - \displaystyle\sum_{n=1}^{\infty} b_nx^n \\ \\ &= 0 \end{align*}$$
   Subtracting $2y_1'$ to both sides of the resulting equation yields
   $$\begin{align*} b_1 + \displaystyle\sum_{n=1}^{\infty} \left[(n+1)nb_{n+1} + (n+1)b_{n+1} - b_n\right]x^n &= -2\left(1 + x + \dfrac{1}{4}x^2 + \dfrac{1}{36}x^3 + \dfrac{1}{576}x^4 +\cdots\right)' \\ \\ b_1 + \displaystyle\sum_{n=1}^{\infty} \left[(n+1)^2b_{n+1} - b_n\right]x^n &= -2\left(1 + \dfrac{1}{2}x + \dfrac{1}{12}x^2 + \dfrac{1}{144}x^3 +\cdots\right) \\ \end{align*}$$
   $$b_1 + \left(4b_2 - b_1\right)x + \left(9b_3-b_2\right)x^2 + \left(16b_4-b_3\right)x^3 +\cdots = -2 - x - \dfrac{1}{6}x^2 - \dfrac{1}{72}x^3 +\cdots$$
   Since the series on both sides converges absolutely we have
   
   $$\begin{align*} b_1 &= -2 \\ \\ b_2 &= \dfrac{b_1 - 1}{4} = \dfrac{-2-1}{4} = -\dfrac{3}{4} \\ \\ b_3 &= \dfrac{b_2 - \frac{1}{6}}{9} = \dfrac{-\frac{3}{4} - \frac{1}{6}}{9} = -\dfrac{11}{108} \\ \\ b_4 &= \dfrac{b_3 - \frac{1}{72}}{16} = \dfrac{-\frac{11}{108}-\frac{1}{72}}{16} = -\dfrac{25}{3456} \end{align*}$$
   Thus the second fundamental solution is given by
   
   $$y_2(x) = y_1(x)\ln(x) - 2x - \dfrac{3}{4}x^2 - \dfrac{11}{108}x^3 - \dfrac{25}{3456}x^4 +\cdots.$$