What is the solution for y^3-(10x+8)+3xy^2y^'=0 ?

The solution for y^3-(10x+8)+3xy^2y^'=0 is y=\sqrt[3]{8+5x+(c_{1})/x}

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y^3-(10x+8)+3xy^2y^'=0

y^{3} - (10 x + 8) + 3 x y^{2} y^{^{'}} = 0

y = 3 8 + 5 x + \frac{c _{1}}{x}

Solution steps

y^{3} - (10 x + 8) + 3 x y^{2} y^{'} (x) = 0

Solve Bernoulli ODE:

y = 3 8 + 5 x + \frac{c _{1}}{x}

y = 3 8 + 5 x + \frac{c _{1}}{x}

\frac{x - 4}{x + 4}

y = \frac{1}{3 x}

\frac{d}{d x} (sin (x^{3} - x))

2 y = x, x + 1 = (y - 1)^{2}

\frac{d}{d x} (7 - 8 x)

What is the solution for y^3-(10x+8)+3xy^2y^'=0 ?
The solution for y^3-(10x+8)+3xy^2y^'=0 is y=\sqrt[3]{8+5x+(c_{1})/x}