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

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

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y^'=(2xy-y^4)/(3x^2)

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

Solution steps

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

Solve Bernoulli ODE:

f (x) = x (x - 10)

5 y^{^{''}} + 32 y = 0, y (0) = 8, y^{^{'}} (0) = 5

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

\int \frac{3}{x ^{3}} d x

f (t) = cos (5 t) + sin (2 t)

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