Question
Solve the following differential equation:
$4\frac{\text{dy}}{\text{dx}}+\text{8y}=\text{5e}^{-3x}$.

Answer

$\frac{\text{dy}}{\text{dy}}+\text{2y}=\frac{5}{4}\text{e}^{-3x}\Rightarrow\text{P(x) = 2, Q(x)}=\frac{5}{4}\text{e}^{-3x}$
I.F. = $\text{e}^{\int\text{p(x)dx}}=\text{e}^{\text{2x}}$
$\therefore \text{We have y}\cdot\text{e}^{\text{2x}}=\int\frac{5}{4}\text{e}^{\text{-3x}}\cdot\text{ e}^{\text{2x}}\text{dx}$
$\Rightarrow\text{y}\cdot\text{e}^{\text{2x}}=\frac{-5}{4}\text{e}^{\text{-x}}+\text{c }\text{ OR }\text{y}=\frac{-5}{4}\text{e}^{\text{-3x}}+\text{c }\text{ e}^{\text{-2x}}.$

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