Solve the following differential equation: dy dx +2y=xe^ 4x - Vidyadip

Question

Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{xe}^{4\text{x}}$

✓

Answer

We have, $\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{xe}^{4\text{x}}\ \dots(1)$ Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ where $\text{P}=2$
$\text{Q}=\text{xe}^{4\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$Multiplying both sides of (1) by $e^{2x}$, we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\times\text{xe}^{4\text{x}}$$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{xe}^{6\text{x}}$
Integrating both sides with respect to x, we get
$\text{e}^{2\text{x}}\text{y}=\int\text{e}^{6\text{x}}\text{xdx + C}$
$\Rightarrow\ \text{e}^{2\text{x}}\text{y}=\text{x}\int\text{e}^{6\text{x}}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{6\text{x}}\text{dx}\Big]\text{dx + C}$
$\Rightarrow\ \text{e}^{2\text{x}}\text{y}=\frac{\text{x}\text{e}^{6\text{x}}}{6}-\frac{\text{e}^{6\text{x}}}{36}+\text{C}$
$\Rightarrow\ \text{y}=\frac{\text{xe}^{4\text{x}}}{6}-\frac{\text{e}^{4\text{x}}}{36}+\text{Ce}^{-2\text{x}}$ Hence, $\text{y}=\frac{\text{xe}^{4\text{x}}}{6}-\frac{\text{e}^{4\text{x}}}{36}+\text{Ce}^{-2\text{x}}$ is the required solution.

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