What is the solution for (dy)/(dx)= 1/(3x^2y^2) ?

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

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(dy)/(dx)= 1/(3x^2y^2)

\frac{d y}{d x} = \frac{1}{3 x ^{2} y ^{2}}

y = 3 - \frac{1}{x} + c_{1}

Solution steps

\frac{d y}{d x} = \frac{1}{3 x ^{2} y ^{2}}

Solve separable ODE:

y = 3 - \frac{1}{x} + c_{1}

y = 3 - \frac{1}{x} + c_{1}

\frac{d}{d x} ((x - 1)^{2} (x - 2) (x - 4))

t an g e n t y = \frac{2 x}{( x ^{2} + 1 )}

\int \frac{6 - 72 x}{1 - 36 x ^{2}} d x

\frac{d}{d x} (\frac{x}{x + 2})

\frac{d}{d x} (\frac{x + 4}{x - 5})

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