Solve $2xy'' + 5y' + xy = 0$ using Frobenius method? Using the Frobenius method, solve the differential equation $2xy'' + 5y' + xy = 0$. 
I've done most of the work, but when it comes to getting the indicial equation I am getting stuck. When working with the sums, all summations start at $n=0$, and one starts at $n=2$. So, combining the sum gives a sum starting at $n=2$, leaving behind the $n=0$ and $n=1$ terms of the other sums. This leads to two indicial equations, and I don't know if that should be possible or not. Thanks in advance.
 A: Let 
$$y=\sum_{n=0}^\infty a_n x^n$$
so the differential equation gives:
$$2\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+5\sum_{n=1}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^{n+1}=0 $$
and by change of index we find
$$2\sum_{n=1}^\infty n(n+1)a_{n+1}x^{n}+5\sum_{n=0}^\infty (n+1)a_{n+1}x^{n}+\sum_{n=1}^\infty a_{n-1}x^{n}=0 $$
hence the index $n=0$ gives
$$a_1=0$$
and for $n\ge1$ we have
$$(2n+5)(n+1)a_{n+1}+a_{n-1}=0$$
Can you take it from here?
A: In General: A second order ODE has a regular singular point at $0$ iff the equation can be written in the form
$$
               y'' + \left(p_{0}\frac{1}{x}+p_{1}+p_{2}x+\cdots\right)y'+\left(q_{0}\frac{1}{x^{2}}+q_{1}\frac{1}{x}+q_{2}+q_{3}x+\cdots\right)y = 0.
$$
The indicial equation is found by substituting $y=x^{\rho}$ into
$$
                    y'' + \frac{p_{0}}{x}y'+\frac{q_{0}}{x^{2}}y=0.
$$
which leads to
$$
                       (\rho(\rho-1)+p_{0}\rho+q_{0})x^{\rho}=0,\\
                          \rho^{2}+(p_{0}-1)\rho+q_{0}=0.
$$
You seem to be interpreting "indicial equation" in a way that different from the way it is used for the method of Frobenius.
Your Equation: The standard form for your equation is
$$
     y''+\frac{5}{2x}y'+\frac{1}{2}y=0.
$$
So your equation does have a regular singular point at $x=0$ with $p_{0}=5/2$, $q_{0}=0$.
Your indicial equation is $\rho^{2}+(p_{0}-1)\rho+q_{0}=\rho(\rho+p_{0}-1)$, which has roots $\rho=0$ and $\rho=-3/2$. Because the difference of the roots is $0-(-3/2)$, which is not an integer, there are solutions  of the form $\sum_{n=0}^{\infty}a_{n}x^{n}$ and $\sum_{n=0}^{\infty}b_{n}x^{n-3/2}$.
