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