What is the solution for 2xyy^'=x^2-y^2 ?

The solution for 2xyy^'=x^2-y^2 is y=sqrt((x^2)/3+(c_{1))/x},y=-sqrt((x^2)/3+(c_{1))/x}

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2xyy^'=x^2-y^2

2 x y y^{^{'}} = x^{2} - y^{2}

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

Solution steps

2 x y y^{'} (x) = x^{2} - y^{2}

Solve Bernoulli ODE:

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

What is the solution for 2xyy^'=x^2-y^2 ?
The solution for 2xyy^'=x^2-y^2 is y=sqrt((x^2)/3+(c_{1))/x},y=-sqrt((x^2)/3+(c_{1))/x}