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

Question

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

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Answer

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}+\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=1+\frac{1}{\text{x}}\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}=1\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\frac{1}{\text{x}}$
$\text{Q}=1$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log\text{x}}$
$=\text{e}^{\log\frac{1}{\text{x}}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\times1$
$\Rightarrow\ \frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\frac{1}{\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\frac{1}{\text{x}}=\int\frac{1}{\text{x}}\text{dx + C}$
$\Rightarrow\ \frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{C}$
Hence, $\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{C}$ is the required solution.

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