Question
Find the general solution of the differential equation
$\text{x}\log\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}=\frac{2}{\text{x}}\cdot\log\text{x}$.

Answer

The given differential equation can be written as
$\frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x log x}}\text{y}=\frac{2}{\text{x}^{2}}$
$\text{I.F.}=\text{e}^{\int\frac{1}{\text{x log x}}\text{dx}}=\text{e}^{\log(\log x)}=\log\text{x}$
The solution is $y . \log x = \int\frac{2}{\text{x}^{2}}\cdot\log\text{x dx + c}$
OR, $y . \log x = 2$
$\Bigg[\log\text{x}\cdot\Big(\frac{-1}{\text{x}}\Big)+\int\frac{\text{dx}}{{\text{x}}^{2}}\Bigg]+\text{c}=2\Bigg[\frac{-\log\text{x}}{\text{x}}-\frac{\text{1}}{\text{x}}\Bigg]+\text{c}$
$\Rightarrow\text{y}\cdot\log\text{x}=-\frac{2}{\text{x}}[1+\log\text{x}]+\text{c}$

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