For y=xe^x, which of the following is correct ?

The correct answer is Option (1) → $x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$

$y=xe^x$ so $\frac{dy}{dx}=xe^x+e^x$

or $\frac{dy}{dx}=y+e^x$

so $\frac{d^2y}{dx^2}=\frac{dy}{dx}+e^x$ [we know that $\frac{y}{x}=e^x$]

so $\frac{d^2y}{dx^2}=\frac{dy}{dx}+\frac{y}{x}$

or $x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$