Apply the Frobenius theorem to a problem Given
$x^2y''+x(x-3)y'+3y=0$
In standard form, we have 
$y''+\frac{x-3}{x}y'+\frac{3}{x^2}y=0$
So $p(x)=\frac{x-3}{x}$ and $q(x)=\frac{3}{x^2}$
Find the first and second solutions. I'm not sure how to start this.. I know I have to find the indicial roots and indicial equation.. but I don't know how. 

Note that your differential equation is of the form
$$x^2y^{\prime\prime} + xp(x)y^{\prime}+q(x)y=0$$
So it follows that $p(x) = x-3$ and $q(x) = 3$ (the important thing to remember in these types of problems is that $p(x)$ and $q(x)$ are polynomials).  Hence $p_0 = p(0) = -3$ and $q_0=q(0)=3$; thus the indicial equation is
$$\begin{aligned} r(r-1)+p_0r+q_0=0 &\implies r(r-1)-3r+3=0\\ &\implies r^2-4r+3=0\\ &\implies r_1=3\text{ and }r_2=1\end{aligned}$$
Now, suppose that the equation $$x^2y^{\prime\prime} +x(x-3)y^{\prime} + 3y = 0\tag{1}$$ has a solution of the form $\displaystyle y=x^r\sum_{n=0}^{\infty}c_nx^n = \sum_{n=0}^{\infty} c_n x^{n+r}$ where $c_0\neq 0$.  It then follows that
$$y^{\prime} = \sum_{n=0}^{\infty}c_n(r+n)x^{r+n-1}\quad\text{and}\quad y^{\prime\prime} = \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}$$
and thus substituting this into $(1)$ gives that
$$\begin{aligned}\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r} +& \sum_{n=0}^{\infty}c_n(n+r)x^{n+r+1}\\ -& 3\sum_{n=0}^{\infty}c_n(n+r)x^{n+r} + 3\sum_{n=0}^{\infty}c_nx^{n+r}=0 \\ \implies \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r} +& \sum_{n=1}^{\infty}c_{n-1}(n+r-1)x^{n+r}\\ -& 3\sum_{n=0}^{\infty}c_n(n+r)x^{n+r} + 3\sum_{n=0}^{\infty}c_nx^{n+r}=0\end{aligned}$$
after shifting the index on the second summation.
If $n=0$, then we get that $$(r(r-1) -3r+3)c_0 = 0$$ and since $c_0\neq 0$ by assumption, this yields the indicial equation we found previously (so we don't get anything new here).  Now for $n\geq 1$, we have that
$$[(n+r)(n+r-1)-3(n+r)+3]c_n+c_{n-1}(n+r-1) =0\tag{2}$$
Note that $-3(n+r) +3 = -3(n+r-1)$ and thus $(2)$ becomes
$$[(n+r)(n+r-1)-3(n+r-1)]c_n+c_{n-1}(n+r-1) =0$$
$$\implies (n+r-3)(n+r-1)c_n = -c_{n-1}(n+r-1)$$
Therefore,
$$c_n = -\frac{c_{n-1}}{n+r-3}\tag{3}$$
for $n+r-1\neq 0$.
We now consider two different cases for solving this recurrence relation.

Case I: $r=r_1=3$.  The recurrence relation $(3)$ simplifies to
$$c_n = -\frac{c_{n-1}}{n}$$
which has the solution $c_n = \dfrac{(-1)^nc_0}{n!}$ (Left for you to verify; should be pretty straightforward).
Therefore, the first Frobenius solution is 
$$y_1(x) = c_0x^{3}\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{n!} = c_0x^3e^{-x}.$$

Case II: $r=r_2=1$.  If we take $r=1$ and $n=2$ in $(2)$, we get 
$$0\cdot c_2+2c_1=0 \implies 0\cdot c_2 - 2c_0 = 0$$
and since $c_0\neq 0$, there is no way to chose $c_2$ such that this equation holds.  Hence, there is no Frobenius solution corresponding to the root $r_2=1$.

Since we only found one Frobenius solution, we'll need to use reduction of order to find the second solution; in particular, if $y_1$ is a solution to the ODE
$$y^{\prime\prime}+P(x)y^{\prime}+Q(x)y = 0,$$
then the second solution is given by
$$y_2 = y_1\int\frac{\exp\left(-\int P(x)\,dx\right)}{y_1^2}\,dx$$
I assume you can take things from here?
