What is the solution for xy^'sin^2(y/x)=x+ysin^2(y/x) ?

The solution for xy^'sin^2(y/x)=x+ysin^2(y/x) is y=-\W(-1/(c_{1)x})x

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xy^'sin^2(y/x)=x+ysin^2(y/x)

x y^{^{'}} sin^{2} (\frac{y}{x}) = x + y sin^{2} (\frac{y}{x})

y = - W (- \frac{1}{c _{1} x}) x

Solution steps

x y^{'} (x) sin^{2} (\frac{y}{x}) = x + y sin^{2} (\frac{y}{x})

Solve using substitution:

y = - W (- \frac{1}{c _{1} x}) x

y = - W (- \frac{1}{c _{1} x}) x

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What is the solution for xy^'sin^2(y/x)=x+ysin^2(y/x) ?
The solution for xy^'sin^2(y/x)=x+ysin^2(y/x) is y=-\W(-1/(c_{1)x})x