Answer
$y=C_{0} e^{x}$
Work Step by Step
We have $y=\Sigma^\infty_{n=1}C_{n} x^{n}$ and
$y'=\Sigma^\infty_{n=0}C_{n+1} x^n (n+1) \times x^{n}$
The equation $y' - y = 0$ becomes :
r, $\Sigma^\infty_{n=0}[C_{n+1}(n+1) - C_{n}]=0$
$C_{1} = C_{0}(n=0)\\C_{2} = \dfrac{C_{0}}{2}=0; (n=1)$;
$C_{3} = (\dfrac{1}{3}) (\dfrac{1}{2}) \times C_{0}=0 ; (n=2) $
This gives: $C_{n} = \dfrac{C_0}{n!}$
Use Formula: $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
Hence, we have $y=C_{0} e^{x}$
