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 $$\frac{y}{x}=t$$

$$\begin{aligned} & \cos \mathrm{t}\left(\frac{\mathrm{dt}}{\mathrm{dx}}\right)=\frac{1}{\mathrm{x}} \\ & \cos \mathrm{t~dt}=\frac{1}{\mathrm{x}} \mathrm{dx} \end{aligned}$$

Integrating both sides

$$\begin{aligned} & \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} \\ & \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c} \end{aligned}$$

Using $$\mathrm{y}(1)=\frac{\pi}{3}$$, we get $$\mathrm{c}=\frac{\sqrt{3}}{2}$$

So, $$\alpha=\sqrt{3} \Rightarrow \alpha^2=3$$