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Question:

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

Solution:

The given differential equation is:

$x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$

This equation is the form of a linear differential equation as:

$\frac{d y}{d x}+p y=Q$ ( where $p=\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ )

Now, I.F $=e^{\int \mu d x}=e^{\int \frac{1}{x \log } d x}=e^{\log (\log x)}=\log x$.

The general solution of the given differential equation is given by the relation,

$y($ I.F. $)=\int(\mathrm{Q} \times \mathrm{I.F} .) d x+\mathrm{C}$

$\Rightarrow y \log x=\int\left(\frac{2}{x^{2}} \log x\right) d x+\mathrm{C}$              …(1)

Now, $\int\left(\frac{2}{x^{2}} \log x\right) d x=2 \int\left(\log x \cdot \frac{1}{x^{2}}\right) d x$.

$=2\left[\log x \cdot \int \frac{1}{x^{2}} d x-\int\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{x^{2}} d x\right\} d x\right]$

$=2\left[\log x\left(-\frac{1}{x}\right)-\int\left(\frac{1}{x} \cdot\left(-\frac{1}{x}\right)\right) d x\right]$

$=2\left[-\frac{\log x}{x}+\int \frac{1}{x^{2}} d x\right]$

$=2\left[-\frac{\log x}{x}-\frac{1}{x}\right]$

$=-\frac{2}{x}(1+\log x)$

Substituting the value of $\int\left(\frac{2}{x^{2}} \log x\right) d x$ in equation (1), we get:

$y \log x=-\frac{2}{x}(1+\log x)+\mathrm{C}$

This is the required general solution of the given differential equation.