The solution of $\frac{dy}{dx} + y = e^{-x}, y(0) = 0$ is

$y = x e^{-x}$

The correct answer is Option (4) → $y = x e^{-x}$ ##

Given that, $\frac{dy}{dx} + y = e^{-x}$

which is a linear differential equation comparing with $\frac{dy}{dx} + P \cdot y = Q$

Here, $P = 1$ and $Q = e^{-x}$

$\text{I.F} = e^{\int 1 \, dx} = e^x$

The general solution is $y \cdot \text{I.F} = \int Q \cdot \text{I.F} \, dx + C$

$y \cdot e^x = \int e^{-x} \cdot e^x \, dx + C$

$\Rightarrow y \cdot e^x = \int dx + C$

$\Rightarrow y \cdot e^x = x + C \dots(i)$

When $x = 0$ and $y = 0$ then, $0 = 0 + C \Rightarrow C = 0$

Eq. (i) becomes $y \cdot e^x = x \Rightarrow y = x \cdot e^{-x}$