How can I find the indicial equation of $x(x-1)y''+3y'-2y=0$? I tried the method of Frobenius but I keep getting lost in the algebra. Is there any other way to get the indicial equation?
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$\begingroup$$y''+\dfrac{3}{x(x-1)}y'-\dfrac{2}{x(x-1)}y=0$ then $p(x)=\dfrac{3}{x(x-1)}$ and $q(x)=-\dfrac{2}{x(x-1)}$. The equation has two regular singular points $x=0$ and $x=1$. For $x=0$ we see $$p_0=\lim_{x\to0}xp(x)=\lim_{x\to0}\dfrac{3}{x-1}=-3$$ and $$q_0=\lim_{x\to0}x^2q(x)=\lim_{x\to0}\dfrac{-2x}{x-1}=0$$ then the indicial equation is $r(r-1)+p_0r+q_0=0$ or $r^2-4r=0$ shows $r=0$ and $r=4$. In this case $4-0\in\mathbb{Z}$ so $r=4$ gives a solution and let $y=x^4(a_0+a_1x+a_2x^2+\cdots)$ to find the answer. Do like this with point $x=1$.
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