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

Question

Solve the following differential equation:

$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$

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Answer

Here, $\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\text{e}^{\text{x}}$

$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\text{e}^{\text{x}}$

It is a linear differential equation, comparing it with

$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$

$\text{P}=\frac{1}{\text{x}}, \text{Q}=\text{e}^{\text{x}}$

I.F. $=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$

$=\text{e}^{\log\text{x}}$

$=\text{x}$

Solution of the equation is given by,

$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$

$\text{y}\times(\text{x})=\int\text{e}^{\text{x}}\times\text{xdx + C}$

$\text{xy}=\text{x}\int\text{e}^{\text{x}}\text{dx}-\int(1\times\int\text{e}^{\text{x}}\text{dx})\text{dx + C}$

Using integration by parts

$=\text{x}\text{e}^{\text{x}}-\int\text{e}^{\text{x}}\text{dx}+\text{C}$

$=\text{x}\text{e}^{\text{x}}-\text{e}^{\text{x}}+\text{C}$

$\text{xy}=(\text{x}-1)\text{e}^{\text{x}}+\text{C}$

$\text{y}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}+\frac{\text{C}}{\text{x}},\text{x}>0$

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