Question
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}$

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=6\text{e}^{\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=2$
$\text{Q}=6\text{e}^{\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=6\text{e}^{2\text{x}}\text{e}^\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=6\text{e}^{3\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=6\int\text{e}^{3\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=6\frac{\text{e}^{3\text{x}}}3+\text{C}$
$\Rightarrow\ ​\text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$
Hence, $\text{ye}^{2\text{x}}=2\text{e}^{3\text{x}}+\text{C}$ is the required solution.

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