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

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

Hint              : You must know the rules of solving differential equation and integration

Given           :  $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$

Solution       :  given differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$

$\frac{d y}{d x}=\frac{y \cos \frac{y}{x}+x}{x \cos \frac{y}{x}} \quad \ldots .(i)$

It is a homogeneous differential equation

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Eqn (i) becomes

\begin{aligned} v+x \frac{d v}{d x} &=\frac{v x \cos v+x}{x \cos v} \\ x \frac{d v}{d x} &=\frac{v \cos v+1}{\cos v}-v \\ x \frac{d v}{d x} &=\frac{v \cos v+1-v \cos v}{\cos v} \end{aligned}

\begin{aligned} &x \frac{d v}{d x}=\frac{1}{\cos v} \\ &\cos v d v=\frac{d x}{x} \end{aligned}

Integrating both sides

\begin{aligned} &\sin v=\log |x|+c \\ &\sin \frac{y}{x}=\log |x|+c \end{aligned}