If sin left(frac{y}{x}right)=log |x|+frac{alp | vedclass

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(A) Given differential equation: $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$.Divide by $x \cos \left(\frac{y}{

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$3$

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(A) Given differential equation: $x \cos \left(\frac{y}{x}ight) \frac{d y}{d x}=y \cos \left(\frac{y}{x}ight)+x$.Divide by $x \cos \left(\frac{y}{x}ight)$ (assuming $x 
eq 0$ and $\cos \left(\frac{y}{x}ight) 
eq 0$):$\frac{d y}{d x} = \frac{y}{x} + \frac{1}{\cos \left(\frac{y}{x}ight)}$.Let $v = \frac{y}{x}$,then $y = vx$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$.Substituting these into the equation:$v + x \frac{d v}{d x} = v + \frac{1}{\cos v}$.$x \frac{d v}{d x} = \sec v$.Separating variables:$\cos v \, dv = \frac{1}{x} \, dx$.Integrating both sides:$\int \cos v \, dv = \int \frac{1}{x} \, dx$.$\sin v = \log |x| + C$.Substituting $v = \frac{y}{x}$ back:$\sin \left(\frac{y}{x}ight) = \log |x| + C$.Given $y(1) = \frac{\pi}{3}$,so $\sin \left(\frac{\pi/3}{1}ight) = \log |1| + C$.$\sin \left(\frac{\pi}{3}ight) = 0 + C \implies C = \frac{\sqrt{3}}{2}$.The solution is $\sin \left(\frac{y}{x}ight) = \log |x| + \frac{\sqrt{3}}{2}$.Comparing this with $\sin \left(\frac{y}{x}ight) = \log |x| + \frac{\alpha}{2}$,we get $\frac{\alpha}{2} = \frac{\sqrt{3}}{2} \implies \alpha = \sqrt{3}$.Therefore,$\alpha^2 = (\sqrt{3})^2 = 3$.

If $\sin \left(\frac{y}{x}\right)=\log |x|+\frac{\alpha}{2}$ 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

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