What is the solution for (dy)/(dx)=(x^3+y^3)/(3xy^2) ?

The solution for (dy)/(dx)=(x^3+y^3)/(3xy^2) is y=\sqrt[3]{(x^3)/2+c_{1}x}

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(dy)/(dx)=(x^3+y^3)/(3xy^2)

\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{3 x y ^{2}}

y = 3 \frac{x ^{3}}{2} + c_{1} x

Solution steps

\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{3 x y ^{2}}

Solve Bernoulli ODE:

y = 3 \frac{x ^{3}}{2} + c_{1} x

y = 3 \frac{x ^{3}}{2} + c_{1} x

f (x) = \frac{2}{x}, at x = 9

(3.2) (- 1.2)

n = 1 \sum \infty \frac{n + 1}{n ^{2}}

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

\int_{0}^{1} 9 x e^{- x^{2}} d x

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