Question
Find the general solution of $\frac{d y}{d x}+3 y=e^{-2 x}$

Answer

It is given that $\frac{d y}{d x}+3 y=e^{-2 x}$ 
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, $p = 3$ and $Q = e^{-2x})$
Now, $I.F. = e^{\int p d x}=e^{\int 3 d x}=e^{3 x}$ 
Thus, the solution of the given differential equation is given by the relation:
$y(\mathrm{I} . \mathrm{F})=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int\left(\mathrm{e}^{-2 \mathrm{x}} \times \mathrm{e}^{3 \mathrm{x}}\right) \mathrm{d} \mathrm{x}+\mathrm{C}$ 
$\Rightarrow \mathrm{ye}^{3 \mathrm{x}}=\int \mathrm{e}^{\mathrm{x}} \mathrm{dx}+\mathrm{C}$ 
$\Rightarrow ye^{3x} = e^x + C$
$\Rightarrow y = e^{-2x} + Ce^{-3x}$
Therefore, the required general solution of the given differential equation is $y = e^{-2x} + Ce^{-3x} $

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