What is the general solution of $\frac{dy}{dx} + y = e^{-x}$?

Explanation

1. 01

   The factor is $e^{x}$: $\frac{d}{dx}(ye^{x}) = e^{x} \cdot e^{-x} = 1$.

2. 02

   $ye^{x} = x + C$, so $y = (x + C)e^{-x}$.