$ 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$
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 $\frac{y}{x}=t$
$\cos t\left(\frac{d t}{d x}\right)=\frac{1}{x}$
$\cos \mathrm{t} \mathrm{dt}=\frac{1}{\mathrm{x}} \mathrm{dx}$
Integrating both sides
$ \sin \mathrm{t}=\ln |\mathrm{x}|+\mathrm{c} $
$ \sin \frac{\mathrm{y}}{\mathrm{x}}=\ln |\mathrm{x}|+\mathrm{c}$
Using $y(1)=\frac{\pi}{3}$, we get $c=\frac{\sqrt{3}}{2}$
So, $\alpha=\sqrt{3} \Rightarrow \alpha^2=3$
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