# DE: $x \, dx + [ sin^2 (y/x) ](y \, dx - x \, dy) = 0$

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Sydney Sales
DE: $x \, dx + [ sin^2 (y/x) ](y \, dx - x \, dy) = 0$

xdx + sin^2 ( y/x ) [ ydx - xdy ] = 0

Jhun Vert

$x\,dx + \left[ \sin^2 \left( \dfrac{y}{x} \right) \right](y\,dx - x\,dy) = 0$

$\dfrac{x\,dx}{x^2} + \left[ \sin^2 \left( \dfrac{y}{x} \right) \right]\left( \dfrac{y\,dx - x\,dy}{x^2} \right) = 0$

$\dfrac{dx}{x} - \left[ \sin^2 \left( \dfrac{y}{x} \right) \right]\left( \dfrac{x\,dy - y\,dx}{x^2} \right) = 0$

$\dfrac{dx}{x} - \left[ \sin^2 \left( \dfrac{y}{x} \right) \right] \, d\left( \dfrac{y}{x} \right) = 0$

$\dfrac{dx}{x} - \dfrac{1}{2}\left[ 1 - \cos \left( \dfrac{2y}{x} \right) \right] \, d\left( \dfrac{y}{x} \right) = 0$

$\dfrac{dx}{x} - \dfrac{1}{2} d\left( \dfrac{y}{x} \right) + \dfrac{1}{2} \cos \left( \dfrac{2y}{x} \right) d\left( \dfrac{y}{x} \right) = 0$

$\dfrac{dx}{x} - \dfrac{1}{2} d\left( \dfrac{y}{x} \right) + \dfrac{1}{4} \cos \left( \dfrac{2y}{x} \right) \left[ d\left( \dfrac{2y}{x} \right) \right] = 0$

$\displaystyle \int \dfrac{dx}{x} - \dfrac{1}{2} \int d\left( \dfrac{y}{x} \right) + \dfrac{1}{4} \int \cos \left( \dfrac{2y}{x} \right) \left[ d\left( \dfrac{2y}{x} \right) \right] = 0$

$\ln x - \dfrac{1}{2} \left( \dfrac{y}{x} \right) + \dfrac{1}{4} \sin \left( \dfrac{2y}{x} \right) = c$           answer

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