What is the solution for 2xy^'-y^3=0 ?

The solution for 2xy^'-y^3=0 is y=sqrt(-1/(ln(x)+c_{1))},y=-sqrt(-1/(ln(x)+c_{1))}

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2xy^'-y^3=0

2 x y^{^{'}} - y^{3} = 0

y = - \frac{1}{ln ( x ) + c _{1}}, y = - - \frac{1}{ln ( x ) + c _{1}}

Solution steps

2 x y^{'} (x) - y^{3} = 0

Solve separable ODE:

y = - \frac{1}{ln ( x ) + c _{1}}, y = - - \frac{1}{ln ( x ) + c _{1}}

y = - \frac{1}{ln ( x ) + c _{1}}, y = - - \frac{1}{ln ( x ) + c _{1}}

\frac{d y}{d x} y = - 3 x^{7} + \frac{3}{5} x^{4} + 568 x + 12

y y^{^{'}} = x e^{x^{2} + y}

y^{^{'}} = (1 - y^{2})^{0.5}

y^{^{'}} = y^{2} (t)

\frac{d y}{d x} x = \frac{2}{3} (y - 1)^{\frac{3}{2}}

What is the solution for 2xy^'-y^3=0 ?
The solution for 2xy^'-y^3=0 is y=sqrt(-1/(ln(x)+c_{1))},y=-sqrt(-1/(ln(x)+c_{1))}