$2^{nd}$ order ODE $4x(1-x)y''-y=0$ with $y'(0)=1$ at $x=0$ it has two singular point $x_0 =1$ and $x_0=1$
I rearrange the equation to 
$$y'' - \frac{1}{4x(1-x)}y =0$$
and by $v(x)=\frac{(x-x_0)^2}{4x(1-x)}$ are analytic at $x_0$ for both $0$ and $1$;
that is, $x_0=1,0$ are regular singular points,
so that, I can find a solution by series... (Here, what kind of method should I apply..?)
I just tried the Method of Frobenius:
where the indicial equation is $$r(r-1)-1=0$$ and $r_1=\frac{1+\sqrt5}{2}$and $r_2=\frac{1-\sqrt5}{2}$ where $r_1 - r_2 =\sqrt5$ is not ingeter.
I require to use $$y_1 = x^\frac{1+\sqrt5}{2} \sum_{n=0}^{\infty}a_n x^n$$
$$y_2 = x^\frac{1-\sqrt5}{2} \sum_{n=0}^{\infty}b_n x^n$$
it seems not sensible to solve this with $x^\frac{1-\sqrt5}{2}$ trem,
should I keep working with Method of Frobenius...?... or in any other way..?

$$a_n=\frac{(4*2*1+1)(4*3*2+1)...(4(n-1)(n-2)+1)}{4^{n-1}n!(n-1)!}=\frac{4^{n-2}(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4^{n-1}n!(n-1)!}=\frac{(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})}{4*n!(n-1)!}$$
Thanks, I just tried it again..for this part
$$I=(2*1+\frac{1}{4})(3*2+\frac{1}{4})...((n-1)(n-2)+\frac{1}{4})$$
$$I=\prod_{n=3}^{+\infty}[(n-1)(n-2)+\frac{1}{4}]$$$$=\prod_{n=3}^{+\infty}[n^2 - 3n +\frac{9}{4}]$$$$=\prod_{n=3}^{+\infty}[(n-\frac{3}{2})^2]$$
$$I=\Gamma^2(n-\frac{3}{2}+1)=\Gamma^2(n-\frac{1}{2})$$
substitute this back, there is $$a_n=\frac{\Gamma^2(n-\frac{1}{2})}{4*\Gamma(n+1)\Gamma(n)}$$
where has a difference of $\pi$ and $4$ of denominator..
b.tw..the computer also gives $\pi$... how does this comes?
