Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (xi) textbook solution.

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

Hint :  you integrate by integrating $x^{n}$

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

Solution : $\text { put in form } \frac{d y}{d x}+P y=Q$

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

Divide bot side by $(x \log x)$

\begin{aligned} &\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \log x \times \frac{1}{x \log x}\\ &\frac{d y}{d x}+\left(\frac{1}{x \log x}\right) y=\frac{2}{x^{2}} \end{aligned}               ...(i)

By comparing (i) with $\frac{d y}{d x}+P y=Q$

\begin{aligned} &P=\frac{1}{x \log x} \quad \text { and } Q=\frac{2}{x^{2}} \\ &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=\int e^{\frac{1}{x \log x} d x} \\ &\text { Let } t=\log x \end{aligned}

\begin{aligned} &d t=\frac{1}{x} d x \\ &I . F=e^{\int \frac{1}{t} d t} \\ &I \cdot F=e^{\log |t|} \\ &\text { I.F }=|t|=\log |x| \end{aligned}

$\begin{gathered} y \times \text { I.F }=\int Q \times \text { I.F } d x+c \\ y \times \log x=\int \frac{2}{x^{2}} \log x d x+c \end{gathered}$                                ....(ii)

$\text { Let } I=2 \int \log x \cdot x^{-2} d x$

\begin{aligned} &I=2\left[\log x \int x^{-2} d x-\int\left\{\frac{1}{x} \int x^{-2} d x\right\} d x\right] \\ &I=2\left[\log x \frac{x^{-1}}{-1}-\int \frac{1}{x} \frac{x^{-1}}{-1} d x\right] \end{aligned}

\begin{aligned} &=2\left(-\log x \frac{1}{x}+\int \frac{1}{x^{2}} d x\right) \\ &I=2\left(\frac{-1}{x} \log x-\frac{1}{x}\right)=\frac{-2}{x}(1+\log x) \end{aligned}

Now enq (ii) becomes

\begin{aligned} &y \log x=I+C \\ &y \log x=\frac{-2}{x}(1+\log x)+c \end{aligned}