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

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

which is a linear differential equation.

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

$IF = {e^{\int d x}} = {e^x}$

The general Solution is
$y \cdot {e^x} = \int {{e^{ - x}}} \cdot {e^x}dx + C$

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

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

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