Math 555: Differential Equations

5.4 Regular Singular Points, Part II

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5.4.6 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

$$xp(x) =\color{#307fe2}{\sum_{n=0}^{\infty} p_nx^n }\qquad\qquad\qquad x^2 q(x) =\color{#ec008c}{\sum_{n=0}^{\infty} q_nx^n } $$
of the form

$$y(x) = \phi(r,x) = \color{#d17300}{\sum_{n=0}^{\infty} a_nx^{n+r}}.$$
Substituting these expressions into the differential equation we obtain

$$\begin{align*} x^2\color{#9e292b}{\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r-2}} + x &\left(\color{#307fe2}{\sum_{n=0}^{\infty} p_nx^n}\right)\left(\color{teal}{\sum_{n=0}^{\infty} a_n(n+r)x^{n+r-1}}\right) +\ldots \\ \\ &\left(\color{#ec008c}{\sum_{n=0}^{\infty} q_nx^n}\right)\left(\color{#d17300}{\sum_{n=0}^{\infty} a_nx^{n+r}}\right) = 0 \end{align*}$$
Distributing the variable $x$ across the absolutely convergent series results in

$$\begin{align*} \color{#9e292b}{\sum_{n=0}^{\infty} a_n(n+r)(n+r-1)x^{n+r}} + &\left(\color{#307fe2}{\sum_{n=0}^{\infty} p_nx^n}\right)\left(\color{teal}{\sum_{n=0}^{\infty} a_n(n+r)x^{n+r}}\right) +\ldots \\ \\ &\left(\color{#ec008c}{\sum_{n=0}^{\infty} q_nx^n}\right)\left(\color{#d17300}{\sum_{n=0}^{\infty} a_nx^{n+r}}\right) = 0 \end{align*}$$
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

$$\begin{align*} L[y] &= L[\phi(r,x)] \\ \\ &= \left(\color{#9e292b}{a_0r(r-1)x^r + a_1(r+1)rx^{r+1}+\cdots+a_n(n+r)(n+r-1)x^{n+r-2}+\cdots}\right) \\ \\ &\quad + \left(\color{#307fe2}{p_0+p_1x+\cdots+p_nx^n+\cdots}\right)\left(\color{teal}{a_0rx^r + a_1(r+1)x^{r+1}+\cdots+a_n(n+r)x^{n+r}+\cdots}\right) \\ \\ &\quad + \left(\color{#ec008c}{q_0+q_1x+\cdots+q_nx^n+\cdots}\right)\left(\color{#d17300}{a_0x^r+a_1x^{r+1}+\cdots+a_nx^{n+r}+\cdots}\right) \end{align*}$$
If we expand the multiplication and group together terms with the same power of $x$,

$$\begin{align*} L[\phi] &= \left(\color{#9e292b}{a_0r(r-1)x^r} + \color{#307fe2}{p_0}\color{teal}{a_0rx^r} + \color{#ec008c}{q_0}\color{#d17300}{a_0x^r}\right) \\ \\ &\qquad + \left(\color{#9e292b}{a_1(r+1)rx^{r+1}} + \color{#307fe2}{p_0}\color{teal}{a_1(r+1)x^{r+1}} + \color{#307fe2}{p_1x}\color{teal}{a_0rx^r} + \color{#ec008c}{q_0}\color{#d17300}{a_1x^{r+1}} + \color{#ec008c}{q_1x}\color{#d17300}{a_0x^r}\right) \\ \\ &\qquad + \left(\color{#9e292b}{a_2(r+2)(r+1)x^{r+2}} + \color{#307fe2}{p_0}\color{teal}{a_2(r+2)x^{r+2}} + \color{#307fe2}{p_1x}\color{teal}{a_1(r+1)x^{r+1}} + \color{#307fe2}{p_2x^2}\color{teal}{a_0rx^r}\right. \\ \\ &\qquad\qquad + \left.\color{#ec008c}{q_0}\color{#d17300}{a_2x^{r+2}} + \color{#ec008c}{q_1x}\color{#d17300}{a_1x^{r+1}} + \color{#ec008c}{q_2x^2}\color{#d17300}{a_0x^r}\right) \\ &\qquad\ddots \\ &\qquad + \left(\color{#9e292b}{a_n(r+n)(r+n-1)x^{r+n}} + \color{#307fe2}{p_0}\color{teal}{a_n(r+n)x^{r+n}} + \color{#307fe2}{p_1x}\color{teal}{a_{n-1}(r+n-1)x^{r+n-1}} +\cdots \right. \\ \\ &\qquad\qquad \left.+\ \color{#307fe2}{p_nx^n}\color{teal}{a_0rx^r} + \color{#ec008c}{q_0}\color{#d17300}{a_nx^{r+n}} + \color{#ec008c}{q_1x}\color{#d17300}{a_{n-1}x^{r+n-1}} +\cdots+ \color{#ec008c}{q_nx^n}\color{#d17300}{a_0x^r} \right) \\ &\qquad\ddots \end{align*}$$
Now we distribute the $a_n$'s and powers of $x$ to simplify each line,

$$\begin{align*} L[\phi] &= a_0\left(\color{#9e292b}{r(r-1)} + \color{#307fe2}{p_0}\color{teal}{r} + \color{#ec008c}{q_0}\right)x^r \\ \\ &\qquad + \left(a_1\left(\color{#9e292b}{(r+1)r} + \color{#307fe2}{p_0}\color{teal}{(r+1)} + \color{#ec008c}{q_0}\right) + a_0\left(\color{#307fe2}{p_1}\color{teal}{r} + \color{#ec008c}{q_1}\right)\right)x^{r+1} \\ \\ &\qquad + \left(a_2\left(\color{#9e292b}{(r+2)(r+1)} + \color{#307fe2}{p_0}\color{teal}{(r+2)} + \color{#ec008c}{q_0}\right) + a_1\left(\color{#307fe2}{p_1}\color{teal}{(r+1)} + \color{#ec008c}{q_1}\right) + a_0\left(\color{#307fe2}{p_2}\color{teal}{r} + \color{#ec008c}{q_2}\right)\right)x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left(a_n\left(\color{#9e292b}{(r+n)(r+n-1)} + \color{#307fe2}{p_0}\color{teal}{(r+n)} + \color{#ec008c}{q_0}\right) + a_{n-1}\left(\color{#307fe2}{p_1}\color{teal}{(r+n-1)} + \color{#ec008c}{q_1}\right) +\cdots \right. \\ \\ &\qquad\qquad \left.+\ a_0\left(\color{#307fe2}{p_n}\color{teal}{r} + \color{#ec008c}{q_n}\right) \right)x^{r+n} \\ &\qquad\ddots \end{align*}$$
If we define

$$F(r) = \left(\color{#9e292b}{r(r-1)} + \color{#307fe2}{p_0}\color{teal}{r} + \color{#ec008c}{q_0}\right),$$
then we have

$$\begin{align*} L[\phi] &= a_0F(r)x^r + \left(a_1F(r+1) + a_0\left(\color{#307fe2}{p_1}\color{teal}{r} + \color{#ec008c}{q_1}\right)\right)x^{r+1} \\ \\ &\qquad + \left(a_2F(r+2) + a_1\left(\color{#307fe2}{p_1}\color{teal}{(r+1)} + \color{#ec008c}{q_1}\right) + a_0\left(\color{#307fe2}{p_2}\color{teal}{r} + \color{#ec008c}{q_2}\right)\right)x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left(a_nF(r+n) + a_{n-1}\left(\color{#307fe2}{p_1}\color{teal}{(r+n-1)} + \color{#ec008c}{q_1}\right) +\cdots + a_0\left(\color{#307fe2}{p_n}\color{teal}{r} + \color{#ec008c}{q_n}\right) \right)x^{r+n} \\ &\qquad\ddots \end{align*}$$
Now we arrange the terms after $a_nF(r+n)$ to increase from $0$ to $n-1$

$$\begin{align*} L[\phi] &= a_0F(r)x^r + \left(a_1F(r+1) + a_0\left(\color{#307fe2}{p_1}\color{teal}{r} + \color{#ec008c}{q_1}\right)\right)x^{r+1} \\ \\ &\qquad + \left(a_2F(r+2) + a_0\left(\color{#307fe2}{p_2}\color{teal}{r} + \color{#ec008c}{q_2}\right) + a_1\left(\color{#307fe2}{p_1}\color{teal}{(r+1)} + \color{#ec008c}{q_1}\right)\right)x^{r+2} \\ &\qquad\ddots \\ &\qquad + \left(a_nF(r+n) + a_0\left(\color{#307fe2}{p_n}\color{teal}{r} + \color{#ec008c}{q_n}\right) +\cdots + a_{n-1}\left(\color{#307fe2}{p_1}\color{teal}{(r+n-1)} + \color{#ec008c}{q_1}\right)\right)x^{r+n} \\ &\qquad\ddots \\ \end{align*}$$
If we write the differential equation with sum signs instead of ellipses

$$L[\phi] = a_0F(r)x^r + \sum_{n=1}^{\infty}\left( a_nF(r+n) + \sum_{k=0}^{n-1} a_k\left(p_{n-k}(r+k) + q_{n-k}\right) \right)x^{r+n} = 0.$$
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

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

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

Pay attention to the fact that the $\{p_n\}$ and $\{q_n\}$ come from the Taylor series for $xp(x)$ and $x^2q(x)$, respectively. This is how they are found. Also, note that $p_0$ and $q_0$ are the same as the values found when computing

$$ \lim_{x\to 0} xp(x) = p_0\qquad\qquad\qquad \displaystyle\lim_{x\to 0} x^2q(x) = q_0. $$

5.4.7 Statement of the Theorem

Armed with the recurrence relation derived above, it is possible to state a theorem for how to find a series solution at regular singular points for the different cases that arise. The theorem is complicated and has some technical cases, but it is much easier to work directly from the theorem than any other approach. The main thing to keep in mind is to first find the indicial equation and the exponents at the singularity and figure out which case to use from there.

Theorem 5.6.1

Consider 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 both $xp(x)$ and $x^2 q(x)$ have 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 $x=0$.
The second fundamental solution is determined based upon the following cases:

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

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

$$ b_n = \left.\dfrac{d}{dr}a_n\right|_{\,r = r_1} = a_n'(r_1). $$

Case 3

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

$$y_2(x) = ay_1(x)\ln|x| + |x|^{r_2}\left(1 + \displaystyle\sum_{n=1}^{\infty} c_n(r_2)x^n\right)$$
where

$$ c_n = \left.\dfrac{d}{dr}\Big( (r - r_2)a_n(r_2) \Big)\right|_{\,r=r_2} = a_n(r_2) + (r-r_2)a_n'(r_2)$$
(by the product rule).
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 , using reduction of order as we have done in previous examples in this section, or working directly from the formulas above. Each of these series for $y_2$ converges absolutely for at least $|x|<\rho$, but not necessarily at $x=0$.

5.4.8 Practice Using the Theorem

As we can see from statement of Theorem 5.6.1 , the process for finding the first fundamental solution is always the same and only second fundamental solution is case-dependent. Focus on finding the first fundamental solution and know that you will only be asked to find the second fundamental solution in reasonable cases (since this is obviously a lot of work).

Do the exercises here for some practice. These are presented in prose instead of video form since videos with explanations would be very long and difficult to follow.

Exercise 5.4.4

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
   $$p(x) = 2\ \text{ and }\ q(x) = \dfrac{6e^x}{x}.$$
   $$\displaystyle\lim_{x\to 0} xp(x) = \displaystyle\lim_{x\to 0} 2x = 0,$$
   $$\displaystyle\lim_{x\to 0} x^2q(x) = \displaystyle\lim_{x\to 0} 6xe^x = 0.$$
   Therefore $x=0$ is a regular singular point.
   
   
2. Find the exponents at the singular point $x=0$.
   
   View Solution

   
   First, we need to make sure we are thinking of our ODE in the standard form
   
   $$ x^ 2y'' + x\left(xp(x)\right)y' + \left(x^2q(x)\right)y = 0, $$
   so we multiply everything by $x$. Then,
   
   $$\begin{align*} x^2 y'' + 2x^2 y' + 6xe^x y &= 0 \\ \\ x^2 y'' + x\left(2x\right)y' + \left(6xe^x\right)y &= 0. \end{align*} $$
   Hence, the power series for $xp(x)$ and $x^2q(x)$ are
   
   $$\begin{align*} xp(x) &= \sum_{n=0}^\infty p_n x^n \\ \\ &= p_0 + p_1 x + p_2 x^2 + \cdots + p_n x^n + \cdots \\ \\ &= 0 + 2x + 0 + 0 +\cdots+ 0 +\cdots,\\ \\ x^2q(x) &= 6x\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = \sum_{n=0}^{\infty}\dfrac{6}{n!}x^{n+1}\\ \\ &= q_0 + q_1 x + q_2 x^2 +\cdots + q_n x^n + \cdots \\ \\ &= 0 + 6x + 6x^2 + 3x^3 + x^4 +\cdots+ \dfrac{6}{n!}x^{n+1} +\cdots. \end{align*} $$
   Thus $p_0 = q_0 = 0$. The indicial 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 solution 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
   
   $$\begin{align*} y_1(x) &= x^1\left(1 - 4x + \dfrac{17}{3}x^2 - \dfrac{47}{12}x^3 +\cdots \right) \\ &= x + 4x^2 + \dfrac{17}{3}x^3 - \dfrac{47}{12}x^4 +\cdots. \end{align*}$$

Exercise 5.4.5

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 and second 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 5.6.1 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
   
   $$y_2(x) = y_1(x)\ln(x) - 2x - \dfrac{3}{4}x^2 - \dfrac{11}{108}x^3 - \dfrac{25}{3456}x^4 +\cdots.$$

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