#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (v) textbook solution.

Answer : $y=\frac{1}{2} \log x+\frac{c}{\log x}$

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

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

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

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

$\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$

This is a first order linear differential equation of the form

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

The integrating factor $If$  of this differential equation is

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

\begin{aligned} &=e^{\log (\log x)} \; \; \; \; \; \; \; \; \; \quad\left[e^{\log x}=x\right] \\ &=\log x \end{aligned}

Hence, the solution of the differential equation is

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

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