Which relation does $y = xe^x$ satisfy?

Explanation

1. 01

   $\frac{dy}{dx} = (x + 1)e^x$ and $\frac{d^2y}{dx^2} = (x + 2)e^x$.

2. 02

   $(x + 2) - 2(x + 1) + x = 0$, so $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0$. The others leave $4x + 4$, $2$ and $-x$ times $e^x$.