How to solve the Differential Equation $x(1-x)y'' + 2(1-2x)y' -2y = 0$?

Question

this is the indicial Eq. so... $$y=\sum_{m=0}^{\infty} C_mx^{m+r}$$ $$y'=\sum_{m=0}^{\infty} (m+r)C_mx^{m+r-1}$$ $$y''=\sum_{m=0}^{\infty} (m+r)(m+r-1)C_mx^{m+r-2}$$ After these y,y',y" get into eq.$x(1-x)y'' + 2(1-2x)y' -2y = 0$, the indicial eq. would be $ -r(r-1)-4r-c=0$ so answer becomes $r_1=-1,r_2=-2$

$(x-x^2)x^{r-2}\{r(r-1)C_o+(r+1)rC_1x+...\}+(1-2x)x^{r-1}\{rC_o+(r+1)C_1x+...\}-2(C_0+C_1x+C_2x^2+...)=0$

then how do i express $y_1=x^{-1}(c_o+c_1x+c_2x^2+...)=c_0x^{-1}(1+x+x^2...)$ and $y_2$?

Answer 1

Since the indicial roots differ by an integer, you might expect to run into logarithmic terms, but here you turn out to be lucky: the recurrence simplifies to $a_{n+1} = a_n$ and the general solution turns out to be

$$ \frac{c_1}{1-x} + \frac{c_2}{x(1-x)} $$

Answer 2

Hint: One way is grouping term as following $$(x-x^2)y''+(1-2x)y'+(1-2x)y'-2y=0$$ $$\left((x-x^2)y'\right)'+\left((1-2x)y\right)'=0$$ $$\left((x-x^2)y'+(1-2x)y\right)'=0$$ so $$(x-x^2)y'+(1-2x)y=C$$