Finding indicial equation I am given a differential equation:
$$xy'' - 4y' + 5xy = 0$$
I am told that this has a singular point at $x=0$.
I computed:
\begin{align*}x p(x) &= -4\\
x^2 q(x) &= 5x^2
\end{align*}
From this, I deduced that $x=0$ is a regular singular point.
But from here, I am having difficulty though finding the indicial equation in terms of "$r$".
I appreciate any help,
thank you.
 A: Here the equation is  $xy'' - 4y' + 5xy = 0$ . . . . $(1)$
$x=0$ is a regular singular point of the equation $(1)$.
So the equation $(1)$ admits of a Frobenius series of the form $$y(x)=x^r \sum_{n=0}^{\infty} a_n x^n \qquad . . . . . (2)$$ where $a_0\neq 0$ and the series converges for all $x$.
From $(2)$, $$y'(x)=\sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1} \qquad ; \qquad y''(x)= \sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2} \qquad . . . . . (3)$$
Substituting $(2)$ and $(3)$ in $(1)$ we have
$$x\sum_{n=0}^{\infty} (n+r)(n+r-1)a_n x^{n+r-2}-4\sum_{n=0}^{\infty} (n+r)a_n x^{n+r-1}+5x\sum_{n=0}^{\infty} a_n x^{n+r}=0$$ or,
$$\sum_{n=0}^{\infty}a_n (n+r) (n+r-5) x^{n+r-1}+5\sum_{n=0}^{\infty}a_n (n+r) x^{n+r+1}=0 \qquad . . . . . (4)$$
Lowest power of $x$ in equation $(4)$ is $(r-1)$.
Coefficient of $x^{r-1}$ gives the indicial equation $$ a_0r(r-5)=0\implies r^2-5r=0$$ as $a_0 \neq 0$.



Indicial equation: If $x=\alpha$ is a regular singular point of the given differential equation $$u''+P(x)u'+Q(x)u=0$$ then the indicial equation is $$r(r-1)+p_0r+q_0=0$$ where


$$p_0=\lim_{x\to \alpha }(x-\alpha)P(x)$$


$$q_0=\lim_{x\to \alpha} (x-\alpha)^2Q(x)$$


${}$
By this rule we also find the indicial equation as follows:
Here $P(x)=-\frac{4}{x}$ and $Q(x)=5$
So $p_0=-4$ and $q_0=0$ and hence the indicial equation is $r(r-1)+p_0r+q_0=0\implies r(r-1)-4r+0=0\implies r^2-5r=0$
${}$



Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$
where $x \in D \subseteq \mathbb{C}$.


The point $x_0 \in D$ is said to be an ordinary point of the above given differential equation if $P(x)$ and $Q(x)$ are analytic at $x_0$.


If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$, the point $x_0$ is called a singular point of the given differential equation.


A singular point $x_0$ of the given differential equation is said to be regular singular point if the function $(x-x_0)P(x)$ and $(x-x_0)^2 Q(x)$ are analytic at $x_0$ and irregular otherwise.

A: I don't know whether you are taught to study this specific kind of differential equation with polynomial coefficients "out of scratch" or take profit of some already usable results.
Indeed, this differential equation can be casted into a form with generic solutions using combinations of Bessel functions (I see you are aware of that because you have used this tag).
For that, multiply the LHS of your differential equation by variable $x$, then use formulas (3) and (4) of this reference, giving the following general solution :
$$y=x^{5/2}\left[C_1 J_{5/2}(\sqrt{5} x)+C_2 Y_{5/2}(\sqrt{5} x)\right]$$
(parameters values $p=-\dfrac52, q=\dfrac52, a=\sqrt{5}, r=1$ in expression (4)), $C_1, C_2$ being constants given by initial/boundary conditions. 
