Solve ODE via Frobenius Method Solve the ODE 
$$4x^2y'' - 4x^2y' + (1-2x)y = 0.$$
My attempt was using Frobenius' Method. This leads to the indicial equation 
$$r(r-1) + \frac{1}{4} = 0,$$
which has a double root at $r = \frac{1}{2}$ and so we have that 
$$y(x) = \sum ^\infty _{n = 0} a_n x^{n + \frac{1}{2}}.$$
However, upon substituting this into the ODE and doing a bunch of algebra, I obtain the recurrence relation 
$$a_n = \frac{a_{n-1}}{4} .$$
But at the same time, since (again after some manipulation), we have the term 
$$a_0 x^{\frac{1}{2}} - a_1 x^{\frac{3}{2}} - 2 a_0 x^{\frac{3}{2}} = 0 \implies a_0 = a_1 = 0.$$
But, according to Wolfram, the first solution is $c_1 e^x \sqrt{x}$ which is not at all what I obtain. Could someone please do this problem with Frobenius' method and perhaps post what it looks like when done correctly? Thank you!
 A: $$4x^2y'' - 4x^2y' + (1-2x)y = 0.$$
The indicial equation is:
$$r(r-1)+\frac 1 4=0 \implies r=\frac 12$$
Hence,
$$y=\sqrt x S \text { where S is the series } S=\sum_{n=0}^\infty a_nx^n$$
Differentiate:
$$y'=\frac S{2\sqrt x }+ S'\sqrt x$$
$$y''=-\frac S {4x\sqrt x }+\dfrac {S'}{\sqrt x}+ S''\sqrt x$$
Plug this in the original equation and multiply by $\sqrt x$
$$4x^2S''+(4x-4x^2)S'-4xS=0$$
Or more simply:
$$xS''+(1-x)S'-S=0$$
This is Laguerre's differential equation with $\lambda =-1$. 
$$xy''+(1-x)y'+\lambda y=0$$
It's easy to use a simple series  solution to solve this last equation. Note that as expected for $S=e^x \implies S=S'=S''=e^x$ is a solution.
Plug the series $ S=\sum_{n=0}^\infty a_nx^n$ in the last equation you get:
$$-a_0+a_1+\sum_{n=1}^\infty x^n((n+1)^2a_{n+1} -a_n(n+1))=0$$
$$\implies a_0=a_1$$
And the recurrence relation is :
$$a_{n+1}=\dfrac {a_n}{n+1} \implies a_n=\dfrac {a_1}{n!}$$
Therefore
$$\boxed {y_1=\sqrt xS=a_0\sqrt x \sum_{n=0}^\infty \dfrac {x^n}{n!}=a_0\sqrt xe^x }$$
For the second solution since we are in the case of double root in the indicial equation, you have to plug in the equation  the following series:
$$y_2(x)=y_1(x) \ln (x)+\sqrt x \sum_{n=1}^\infty b_nx^n$$
