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

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

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

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

Solution steps

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

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x^{2} y^{^{''}} - 4 x y^{^{'}} + 6 y = 0, y (1) = - 2, y^{^{'}} (1) = 4

\frac{\partial}{\partial x} (3 x^{2} + 2 y^{4})

x \to 0 lim (lo g_{e} (1 + x))

\int sin (x) (1 + sec^{2} (x)) d x

\frac{d x}{d t} = 1.5 t cos^{2} (x), x (0) = - 1.4

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