What is the solution for y^3-(18x+4)+3xy^2y^'=0 ?

The solution for y^3-(18x+4)+3xy^2y^'=0 is y=\sqrt[3]{4+9x+(c_{1})/x}

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y^3-(18x+4)+3xy^2y^'=0

y^{3} - (18 x + 4) + 3 x y^{2} y^{^{'}} = 0

y = 3 4 + 9 x + \frac{c _{1}}{x}

Solution steps

y^{3} - (18 x + 4) + 3 x y^{2} y^{'} (x) = 0

Solve Bernoulli ODE:

y = 3 4 + 9 x + \frac{c _{1}}{x}

y = 3 4 + 9 x + \frac{c _{1}}{x}

\frac{\partial}{\partial x} (e^{- x} \cdot (x - y^{2}))

\frac{d}{d x} (arcsech (1 - x^{2}))

\frac{d}{d x} (x \cdot e^{x y})

\frac{d}{d x} ((cos^{2} (x))^{x})

x \to 0 - lim (x^{3} + 3 x + 2)

What is the solution for y^3-(18x+4)+3xy^2y^'=0 ?
The solution for y^3-(18x+4)+3xy^2y^'=0 is y=\sqrt[3]{4+9x+(c_{1})/x}