What is the solution for y^3-(2x+4)+3xy^2y^'=0 ?

The solution for y^3-(2x+4)+3xy^2y^'=0 is y=\sqrt[3]{4+x+(c_{1})/x}

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y^3-(2x+4)+3xy^2y^'=0

y^{3} - (2 x + 4) + 3 x y^{2} y^{^{'}} = 0

Solution steps

y^{3} - (2 x + 4) + 3 x y^{2} y^{'} (x) = 0

Solve Bernoulli ODE:

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\frac{\partial}{\partial x} (sin (t))

\int \frac{1}{- t ^{2} + 4 t - 3} d t

ln (cos (3 x))

What is the solution for y^3-(2x+4)+3xy^2y^'=0 ?
The solution for y^3-(2x+4)+3xy^2y^'=0 is y=\sqrt[3]{4+x+(c_{1})/x}