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

The solution for y^'=(x^4y^3+2x^4)/(y^2) is y=\sqrt[3]{e^{(3x^5-c_{1})/5}-2}

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

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

y = 3 e^{\frac{3 x ^{5} - c _{1}}{5}} - 2

Solution steps

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

Solve Bernoulli ODE:

y = 3 e^{\frac{3 x ^{5} - c _{1}}{5}} - 2

y = 3 e^{\frac{3 x ^{5} - c _{1}}{5}} - 2

\frac{d}{d x} (- \frac{1}{8} cos (\frac{x}{2}))

\int \frac{x}{6} e^{x^{2}} d x

\frac{\partial}{\partial x} (sin (7 y^{2} - 7 x y))

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

\frac{s + 2}{s ( s ^{2} + 9 )}

What is the solution for y^'=(x^4y^3+2x^4)/(y^2) ?
The solution for y^'=(x^4y^3+2x^4)/(y^2) is y=\sqrt[3]{e^{(3x^5-c_{1})/5}-2}