Question:
$\frac{d y}{d x}+3 y=e^{-2 x}$
Solution:
The given differential equation is $\frac{d y}{d x}+p y=Q$ (where $p=3$ and $Q=e^{-2 x}$ ).
Now, I.F $=e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$.
The solution of the given differential equation is given by the relation,
$y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C}$
$\Rightarrow y e^{3 x}=\int\left(e^{-2 x} \times e^{3 x}\right)+\mathrm{C}$
$\Rightarrow y e^{3 x}=\int e^{x} d x+\mathrm{C}$
$\Rightarrow y e^{3 x}=e^{x}+\mathrm{C}$
$\Rightarrow y=e^{-2 x}+\mathrm{C} e^{-3 x}$
This is the required general solution of the given differential equation.