If (yx)=_e|x|+2 is the solution of x(yx)dydx=y(yx)+x and y(1)=3, then ^2 is equal to:

Question

If $\sin\left(\dfrac{y}{x}\right)=\log_e|x|+\dfrac{\alpha}{2}$ is the solution of $x\cos\left(\dfrac{y}{x}\right)\dfrac{dy}{dx}=y\cos\left(\dfrac{y}{x}\right)+x$ and $y(1)=\dfrac{\pi}{3}$, then $\alpha^2$ is equal to:

Accepted Answer

With $y=vx$: $x\cos v\left(v+x\dfrac{dv}{dx}\right) = vx\cos v+x \implies x\cos v\,\dfrac{dv}{dx}=1 \implies \sin v = \ln|x|+C$.

Using $y(1)=\pi/3 \Rightarrow \sin(\pi/3)=\dfrac{\sqrt3}{2}=C$, so $\alpha=\sqrt3$, $\alpha^2=3$.

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