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

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

Here, $P = 1,Q = {e^{ - x}}$
$\mid F = {e^{\int P dx}} = {e^{\int d x}} = {e^x}$

The general Solution is

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

$\Rightarrow$ $y \cdot {e^x} = \int d x + C$

$\Rightarrow$ $y \cdot {e^x} = x + C$

………(i)
When $x = 0$ and $y = 0$, then
$0 = 0 + C \Rightarrow C = 0$

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

Eq.$\Rightarrow$ $y = x{e^{ - x}}$