TITLE: Behaviour of solutions to ODE near singular points
QUESTION [7 upvotes]: I am having trouble understanding how to classify what happens to solutions of an ODE near singular points. For example;
I have a question that is about the ODE given by;
$$(x^2-36)y''+(6-x)y'+(x^2+12x+36)y=0$$  
And the question asks about the behaviour of the solutions near each of the singular points.
I found that the singular points were $x_1=6$ and $x_2=-6$
And I found that by taking limits, that both point were regular because of the finite limits,
$$\lim_{x \to 6} (x-6)\frac{6-x}{x^2-36}=0$$ and $$\lim_{x \to 6} \frac{(x-6)^{2}(x^2+12x+36}{x^2-36}$$ 0
and similarly
$$\lim_{x \to -6}(x+6)^{2} \frac{6-x}{x^2-36}$$ 0 and  $$\lim_{x \to -6} \frac{(x+6)^{2}(x^2+12x+36)}{x^2-36}=0$$ 
So what can I infer from this at all? I see that it has two real roots, but what can this tell me about the solution's behaviour? Can someone please take some time to help. 
Do I just need to look at the Euler equation corresponding to the case of two real distinct roots?
By the way, I do not think the answer is supposed to take any extreme amounts of work, it is from a past online assignment, i,e 
That is, I have what the answer is already, but I want to understand it. I really do appreciate the answers given but they are definitely beyond my level and beyond what we are even supposed to consider. We only have covered things such as that behaviour of points is similar to the behaviour of the solutions of the associated euler, and things of that level, etc. that is , the answer provided by the user a couple days ago seems to have left the impression that the question is answered but unfortunately it really is of no help to me even though I do appreciate the effort done 
Another example of the type of question is to do the same thing but say for a different form such as ,
$$(x^2-4x-21)^{2}y''+(x^2-9)y'-xy=0$$
For which I found the two singular points to be $x_o=-3$, a regular singular point and $x_1=7$, an irregular singular point. And I am wanting to know the same things about the solutions in regard to $x_o$, for example, how can I know if all solutions remain bounded near it? or if all non zero solutions are unbounded, etc? 
Should I just look at the limits as x goes to the solutions of the euler equation as above? Or is there something else going on too?
 I have been trying to understand it for a while and I just can't seem to figure it out. Thanks a lot 
PS: I am really still trying to figure this out, I have an exam coming up want to learn this.

REPLY [5 votes]: It seems for some reason, you haven't learn the characteristic exponents for a regular singular point.
Consider following $2^{nd}$ order ODE which has a regular singular point at $x = 0$.
$$x^2 y''(x) + xp(x) y'(x) + q(x) y(x) = 0\
\quad\text{ where }\quad
\begin{cases}
p(z) &= \sum\limits_{k=0}^\infty p_k z^k\\
q(z) &= \sum\limits_{k=0}^\infty q_k z^k\\
\end{cases}
\;\;\text{ for } |z| < \rho.
$$
For simplicity, we will assume all coefficients $p_k, q_k$ are real. From these coefficients, we can construct a quadratic equation
$$r(r-1) + p_0 r + q_0 = 0$$
This is known as the indicial equation for the ODE.
Let $r_1, r_2$ be the two roots of indicial equations. They are known as the characteristic exponents and the convention is labeling them to make $\Re(r_1) \ge \Re(r_2)$. 
Using these two exponents, we can say something about the solutions of the ODE near $0$.

If $r_1, r_2$ are complex or $r_1, r_2$ are real but $r_1 - r_2 \notin \mathbb{N}$, then the ODE has two independent solutions of the form:
$$y_1(x) = x^{r_1} \sum_{k=0}^\infty a_k x^k
\quad\text{ and }\quad
y_2(x) = x^{r_2} \sum_{k=0}^\infty b_k x^k$$
Otherwise, $r_1, r_2$ are real but $r_1 - r_2 \in \mathbb{N}$, the two linear independent solutions has the form
$$y_1(x) = x^{r_1} \sum_{k=0}^\infty a_k x^k
\quad\text{ and }\quad
y_2(x) = K y_1(x)\log x + x^{r_2} \sum_{k=0}^\infty b_k x^k$$
If $r_1 = r_2$, $K \ne 0$ and we can choose it to be $1$. If $r_1 > r_2$, $K$ can be zero, can be non-zero.

Apply this to our ODE
$$\begin{align}
& (x^2-36)y''+(6-x)y'+(x^2+12x+36)y=0\\
\iff & y'' - \frac{1}{x+6} y' + \frac{x+6}{x-6} y = 0
\end{align}
$$
It has two regular singular points at $x = -6$ and $6$.

At $x = -6$, the indicial equation is
$$r(r-1) - r = 0 \quad\implies\quad r_1 = 2, r_2 = 0$$
So the two solutions of the ODE have following forms:
$$\left\{\begin{align}
y_1(x) &= (x+6)^2 \sum_{k=0}^\infty a_k (x+6)^k\\
y_2(x) &= K y_1(x)\log(x+6) + \sum_{k=0}^\infty b_k (x+6)^k
\end{align}\right.
$$
Independent of whether $K = 0$ or not, the $\log(x+6)$ term in $y_2(x)$ will be suppressed by the $(x+6)^2$ factor in $y_1(x)$. As a result, both solutions $y_1(x), y_2(x)$ and hence all solutions of the ODE will be bounded near $x = -6$.
At $x = 6$, the indicial equation is
$$r(r-1) = 0 \quad\implies\quad r_1 = 1, r_2 = 0$$
The two solutions there have following forms:
$$\left\{\begin{align}
y_1(x) &= (x-6) \sum_{k=0}^\infty a_k (x-6)^k\\
y_2(x) &= K y_1(x)\log(x-6) + \sum_{k=0}^\infty b_k (x-6)^k
\end{align}\right.
$$
Similar to the $x = -6$ case, the $\log(x-6)$ term in $y_2(x)$ get suppressed by the
$(x-6)$ factor in $y_1(x)$. As a result, all solutions to the ODE is bounded near $x = 6$.

For the other ODE
$$\begin{align}
 &(x^2-4x-21)^{2}y''+(x^2-9)y'-xy=0\\
\iff & y'' + \frac{x-3}{(x-7)^2(x+3)} y' + \frac{x}{(x-7)^2(x+3)^2} y = 0
\end{align}
$$
It has a regular singular point at $x = -3$ and an irregular one at $x = 7$. I have no idea how to deal with the irregular singular point. For the regular singular point at $x = -3$, the corresponding indicial equation is
$$r(r-1) - \frac{3}{50} r -\frac{3}{100} = 0
\quad\implies\quad
\begin{cases}
r_1 &= \frac{53+\sqrt{3109}}{100} \approx +1.087584074378026\\
r_2 &= \frac{53-\sqrt{3109}}{100} \approx -0.027584074378026
\end{cases}
$$
Since $r_1, r_2$ are real but $r_1 - r_2$ isn't an integer, the two solutions
has the from
$$\left\{\begin{align}
y_1(x) &= (x+3)^{r_1} \sum_{k=0}^\infty a_k(x+3)^k\\
y_2(x) &= (x+3)^{r_2} \sum_{k=0}^\infty b_k(x+3)^k
\end{align}\right.
$$
Since $r_1 > 0$ while $r_2 < 0$, $y_1(x)$ is bounded near $x = -3$ while $y_2(x)$ is unbounded there.