Answer
$y=C_{0}\times e^{x}$
Work Step by Step
$y' - y = 0$
By using $y=\Sigma^\infty_{n=1}C_{n}\times x^{n}$
Find the derivative
$y'=\Sigma^\infty_{n=1}C_{n}\times n \times x^{n-1}$
$y'=\Sigma^\infty_{n=0}C_{n+1}\times (n+1) \times x^{n}$
Fill in for the original equation:
$\Sigma^\infty_{n=0}C_{n+1}\times (n+1) \times x^{n}$ - $\Sigma^\infty_{n=1}C_{n}\times x^{n}=0$
Make into one sum
$\Sigma^\infty_{n=0}(C_{n+1}\times (n+1) - C_{n})=0$
Fill some values for $n$ to determine the coefficients.
$n=0 : C_{1} = C_{0}$
$n=1 : C_{2} = \frac{1}{2}C_{0}=0$
$n=2 : C_{3} = \frac{1}{3}\times\frac{1}{2}C_{0}=0$
$C_{n} = \frac{1}{n!}C_{0}$
Notice: $y=\Sigma^\infty_{n=0} \frac{x^n}{n!} \times e^{x}$
$y=C_{0}\Sigma^\infty_{n=0} \frac{1}{n!} \times x^{n}$
$y=C_{0}\times e^{x}$
