Answer
$C_0+C_1 \Sigma^\infty_{n=1} \dfrac{x^n}{n!}=C_0-C_1\ln(1-x)$
Work Step by Step
Given: $y''+xy'+y=0$
Since, $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
Here, we have
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$ and $y''=\Sigma^\infty_{n=2}n(n-1) C_{n} x^{n-2}$
Then, the given equation $(x-1)y''+y'=0$ becomes:
$\Sigma^\infty_{n=2}n(n-1) C_{n} x^{n-2}+\Sigma^\infty_{n=1}n c_n x^{n-1}=0$
or, $\Sigma^\infty_{n=1}n(n-1)c_{n}x^{n-1}-\Sigma^\infty_{n=1}(n+1)n c_{n+1}x^{n-1}+\Sigma^\infty_{n=1}n c_{n}x^{n-1}=0$
or, $c_{n+1}=\dfrac{n}{n+1}c_n$
For n=2 , we have $C_{2} =\dfrac{C_{1}}{2}$;
For n=3 , we have $C_{3} =\dfrac{C_{1}}{3}$;
Thus, we get $C_{n} = \dfrac{1}{n}C_n$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, $y= \dfrac{(-1)^n}{2 \cdot 4 \cdot 6...(2n)}C_0+ \dfrac{(-1)^n}{3 \cdot 5 \cdot 7...(2n+1)}C_1$
or, $y= C_0+C_1 \Sigma^\infty_{n=1} \dfrac{x^n}{n!}=C_0-C_1\ln(1-x)$
