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

The solution for y^'=((y^2+x^2))/(2xy) is y=sqrt(x^2+c_{1)x},y=-sqrt(x^2+c_{1)x}

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

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

y = x^{2} + c_{1} x, y = - x^{2} + c_{1} x

Solution steps

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

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y = x^{2} + c_{1} x, y = - x^{2} + c_{1} x

y = x^{2} + c_{1} x, y = - x^{2} + c_{1} x

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What is the solution for y^'=((y^2+x^2))/(2xy) ?
The solution for y^'=((y^2+x^2))/(2xy) is y=sqrt(x^2+c_{1)x},y=-sqrt(x^2+c_{1)x}