What is the solution for y^'=((x^3y^3+4x^3))/(y^2) ?

The solution for y^'=((x^3y^3+4x^3))/(y^2) is y=\sqrt[3]{e^{(3x^4-c_{1})/4}-4}

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y^'=((x^3y^3+4x^3))/(y^2)

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

Solution steps

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

Solve Bernoulli ODE:

\int_{\frac{1}{5}}^{3} 12 x ln (5 x) d x

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

f (x) = 5 x^{2} - 5 x + 9

\frac{d}{d x} (- 9 x^{3} + 27 x + 7 x - 66)

3 x, at x = 4

What is the solution for y^'=((x^3y^3+4x^3))/(y^2) ?
The solution for y^'=((x^3y^3+4x^3))/(y^2) is y=\sqrt[3]{e^{(3x^4-c_{1})/4}-4}