Can some one show me how you do this kind of problem by Frobenius? I have a homework need to due with Frobenius method. However I was sick on the that we covered that in class so I don't know how to do this problem. Can someone show me all the process the way you do? 
The question is 

$x=0$ is a regular singular point of the given differential equation    $$xy"+(x+1)y'-\frac{y}{9x}=0.$$
  Use the method of Frobenius to obtain two linearly independent series solutions about $x=0$. You must show at least $4$ nonzero terms of each series solutions.

Please help me! 
 A: Your second line $y'' + (x+1)y' - \frac19 y = 0$ does not make sense: it should be $y'' + \frac{x+1}{x} y' - \frac1{9x^2} y = 0$ which is obtained by dividing the first line through by $x$ in order to make the coefficient of $y''$ equal to $1$.  Nevertheless, you did extract the correct functions $p(x)$ and $q(x)$, so it doesn't invalidate the remainder.
You appear to have solved the indicial equation correctly to obtain the roots $r=\pm\frac13$.  Now all that's left to do is to actually look for the series solutions one term at a time.  I'll just work out a couple terms of the case $r=\frac13$ and I think you should have no trouble doing the rest.
We are looking for a solution of the form $$y(x) = a_0 x^{1/3} + a_1 x^{4/3} + a_2 x^{7/3} + \cdots.$$  Substituting this into the original equation (or the second line, once you've corrected it) gives:
$$\begin{align} xy'' = x \big(- \frac29 a_0 x^{-5/3} + \frac49 a_1 x^{-2/3} + \cdots\big) &= -\frac29 a_0 x^{-2/3} + \frac49 a_1 x^{1/3} + \cdots \\
(x+1)y' = (x+1) \big(\frac13 a_0 x^{-2/3} + \frac43 a_1 x^{1/3} + \cdots\big) &= \frac13 a_0 x^{-2/3} + (\frac13 a_0 + \frac43 a_1) x^{1/3} + \cdots \\
-\frac{y}{9x} &= -\frac19 a_0 x^{-2/3} - \frac19 a_1 x^{1/3} + \cdots.
\end{align}$$
Summing these three terms should give a series with all $0$ terms, so we have the constraints:
$$\begin{align} -\frac29 a_0 + \frac13 a_0 - \frac19 a_0 &= 0 \\
  \frac49 a_1 + \frac13 a_0 + \frac43 a_1 - \frac19 a_1 &= 0.\end{align}$$
The first of these simplifies to $0 = 0$, which is as expected since it just verifies that the indicial equation was solved correctly: since this is a linear equation the leading coefficient $a_0$ can typically be arbitrary, since we can multiply everything by a constant. The second constraint simplifies to something more concrete:
$$\frac23 a_0 + \frac43 a_1 = 0 \implies a_1 = -\frac12 a_0.$$
So the series solution for $r=\frac13$ starts as $a_0 \big( x^{1/3} - \frac12  x^{4/3} + \cdots\big) $.
This illustrates the basic idea of Frobenius's method, and it should be clear how you can keep going to higher-order terms to extract the relations between $a_2$ and $a_1$, etc.  For your first attempt I would suggest doing it this long way to understand what's going on.
Once you are confident you can "automate" the process somewhat by computing the derivatives in a more generic way for $a_k x^{1/3 + k}$ and then you can deduce a general equation that relates $a_{k+1}$ to $a_k$, and thereby find the general form of the series.  But since this question only asks for the first 4 terms, you may find it simpler to work with $a_2, a_3, a_4$ explicitly as above.
