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

Question

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

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Answer

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=\text{x}\log\text{x}$
Dividing both sides by x, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\log{\text{x}}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\log\text{x}$
Now,
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{\log|\text{x}|}=\text{x}$
So, the solution is given by
$\text{y}\times\text{I.F.}=\int\text{Q}\times\text{I.F. dx + C}$
$\Rightarrow\ \text{xy}=\int\text{x}\log\text{x dx + C}$
$\Rightarrow\ \text{xy}=\log\text{x}\int\text{xdx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x dx}\Big]\text{ dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^2\log\text{x}}{2}-\int\frac{\text{x}}2\text{dx + C}$
$\Rightarrow\ \text{xy}=\frac{\text{x}^2\log\text{x}}{2}-\frac{\text{x}^2}4+\text{C}$
$\Rightarrow\ 4\text{xy}=2\text{x}^2\log\text{x}-\text{x}^2+\text{K}$ (where, K = 2C)

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