What is the solution for y^'=(-y^3+14x-8)/(3xy^2) ?

The solution for y^'=(-y^3+14x-8)/(3xy^2) is y=\sqrt[3]{-8+7x+(c_{1})/x}

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y^'=(-y^3+14x-8)/(3xy^2)

y^{^{'}} = \frac{- y ^{3} + 14 x - 8}{3 x y ^{2}}

y = 3 - 8 + 7 x + \frac{c _{1}}{x}

Solution steps

y^{'} (x) = \frac{- y ^{3} + 14 x - 8}{3 x y ^{2}}

Solve Bernoulli ODE:

y = 3 - 8 + 7 x + \frac{c _{1}}{x}

y = 3 - 8 + 7 x + \frac{c _{1}}{x}

\frac{d}{d x} (\frac{2}{2 x ^{4} - 5})

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

\int \frac{2 x + 3}{x ^{2} + 2 x + 5} d x

\frac{\partial}{\partial x} (e^{- \frac{y}{x ^{2} + y ^{2} + 1}})

(- 3.7) (0.3)

What is the solution for y^'=(-y^3+14x-8)/(3xy^2) ?
The solution for y^'=(-y^3+14x-8)/(3xy^2) is y=\sqrt[3]{-8+7x+(c_{1})/x}