Solve the following differential equation: dy dx -y=xe^ x

Question

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

✓

Answer

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

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