Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 Subquestion (xi)  maths textbook solution.

Answer : $y=\log x+\frac{c}{\log x}$

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

Hint : Using $\int \frac{1}{x} d x$

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

Divide by $x\; \log\; x$

$=\frac{d y}{d x}+\left(\frac{1}{x \log x}\right) y=\frac{2}{x}$

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=\frac{1}{x \log x} \text { and } Q=\frac{2}{x} \end{aligned}

The integrating factor $If$  of this differential equation is

\begin{aligned} &I f=e^{\int \frac{1}{x \log x} d x} \\ &=e^{\log |\log x|} \\ &=\log x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[e^{\log e^{x}}=x\right] \end{aligned}

Hence, the solution of the different equation is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y \log x=\int\left(\frac{2}{x} \log x\right) d x+C \end{aligned}

\begin{aligned} &=y \log x=2 \int\left(\frac{1}{x} \log x\right) d x+C \\ &=y \log x=\frac{2(\log x)^{2}}{2}+C \quad\left[\int x d x=\frac{x^{2}}{2}+C\right] \\ &=y \log x=(\log x)^{2}+C \end{aligned}

Divide by $\log\; x$, we get

\begin{aligned} &=\frac{y \log x}{\log x}=\frac{(\log x)^{2}}{\log x}+\frac{C}{\log x} \\ &=y=\log x+\frac{C}{\log x} \end{aligned}