$\cfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}$

: The given equation is $\cfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}$

…(1)
which is a linear equation of the type $\cfrac{{dy}}{{dx}} + Py = Q$

Where,$P = 3$and $Q = {e^{ - 2x}}$

$\therefore {\rm{I}}{\rm{.F}}{\rm{.}} = {e^{\int {Pdx} }} = {e^{\int {3dx} }} = {e^{3x}}$

The solution is $y \cdot \left( {{\rm{I}}{\rm{.F}}{\rm{.}}} \right) = \int Q \cdot ({\rm{I}}.{\rm{F}}.)dx + C$

$\Rightarrow y{e^{3x}} = \int {{e^{3x}}} \cdot {e^{ - 2x}}dx + C \Rightarrow y \cdot {e^{3x}} = \int {{e^x}} dx + C$

$\Rightarrow v \cdot {e^{3x}} = {e^x} + C$

$\Rightarrow y = {e^{ - 2x}} + C{e^{ - 3x}}$, which is the required solution.