Question

If $y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+.......\infty$, then show that $\frac{dy}{dx}=y$

Answer 1

$y=1+x+\left(\frac{x^2}{2!}\right)+\left(\frac{x^3}{3!}\right)+...........+\infty$

$\left(\frac{dy}{dx}\right)=0+1+x+\left(\frac{x^2}{2!}\right)+...........+\infty$

$=y$

we can also solve it this way also .

as expansion of $e^x=1+x+\left(\frac{x^2}{2!}\right)+\left(\frac{x^3}{3!}\right)+...........+\infty$

$ory=e^x$

$\therefore \left(\frac{dy}{dx}\right)=e^x$