## Calculus: Early Transcendentals 8th Edition

$y=C_{0} e^{x}$
Given: $y' - y = 0$ Since, $y=\Sigma^\infty_{n=1}C_{n} x^{n}$ Here, we have $y'=\Sigma^\infty_{n=0}C_{n+1}(n+1) x^{n}$ Then, the given equation $y' - y = 0$ becomes: $\Sigma^\infty_{n=0}C_{n+1} (n+1)x^{n}$ - $\Sigma^\infty_{n=1}C_{n}x^{n}=0$ or, $\Sigma^\infty_{n=0}[C_{n+1}(n+1) - C_{n}]=0$ For n=0 , we have $C_{1} = C_{0}$; For n=1 , we have $C_{2} = \dfrac{C_{0}}{2}=0$; For n=2 , we have $C_{3} = (\dfrac{1}{3}) (\dfrac{1}{2}) \times C_{0}=0$ Thus, we get $C_{n} = \dfrac{C_0}{n!}$ We know that $\Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x}$ Hence, $y=C_{0} e^{x}$