If $\sin \left(\frac{y}{x}\right)=\log _e|x|+\frac{\alpha}{2}$ is the solution of the differential equation $$x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$$ and $y(1)=\frac{\pi}{3}$, then $\alpha^2$ is equal to

<p>Differential equation :-</p> <p>$$\begin{aligned} & x \cos \frac{y}{x} \frac{d y}{d x}=y \cos \frac{y}{x}+x \\ & \cos \frac{y}{x}\left[x \frac{d y}{d x}-y\right]=x \end{aligned}$$</p> <p>Divide both sides by $\mathrm{x}^2$</p> <p>$\cos \frac{y}{x}\left(\frac{x \frac{d y}{d x}-y}{x^2}\right)=\frac{1}{x}$</p> <p>Let $\frac{y}{x}=t$</p> <p>$$\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$$</p> <p>Integrating both sides</p> <p>$$\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}$$</p> <p>Using $\mathrm{y}(1)=\frac{\pi}{3}$, we get $\mathrm{c}=\frac{\sqrt{3}}{2}$</p> <p>So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$</p>