Answer
$y=C_0-C_1 \ln (1-x)$
Work Step by Step
We have $\Sigma^\infty_{n=2}n(n-1) C_{n} x^{n-2}+\Sigma^\infty_{n=1}n c_n x^{n-1}=0$
$\implies \Sigma^\infty_{n=1}n(n-1)c_{n}x^{n-1}-\Sigma^\infty_{n=1}(n+1)n c_{n+1}x^{n-1}+\Sigma^\infty_{n=1}n c_{n}x^{n-1}=0$
This gives: $c_{n+1}=\dfrac{n}{n+1}c_n$
$C_{2} =\dfrac{C_{1}}{2}$; when $n=2$
$C_{3} =\dfrac{C_{1}}{3}$; when $n=3$
$\implies C_{n} = \dfrac{1}{n}C_n$
This implies that
$y= \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)} \times C_0+ \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)} \times C_1$
Thus, w ehave $y= C_0+C_1 \Sigma^\infty_{n=1} \dfrac{x^n}{n!}$
or, $y=C_0-C_1 \ln (1-x)$
