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

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

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

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

Solution steps

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

Solve separable ODE:

x^{2} e^{4 x}

\frac{d y}{d x} = e^{- ka \cdot x}, y (0) = 0

\int \frac{x ^{2} - x + 12}{x ^{3} + 6 x} d x

\frac{\partial}{\partial x} (e^{3 x} sin (cy))

n = 0 \sum \infty \frac{3 ^{n}}{2 ^{n}}

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