What is the solution for xy^2(dy)/(dx)=y^3-x^3,y(1)=3 ?

The solution for xy^2(dy)/(dx)=y^3-x^3,y(1)=3 is y=\sqrt[3]{3}x\sqrt[3]{-ln(x)+9}

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xy^2(dy)/(dx)=y^3-x^3,y(1)=3

x y^{2} \frac{d y}{d x} = y^{3} - x^{3}, y (1) = 3

y = 33 x 3 - ln (x) + 9

Solution steps

x y^{2} \frac{d y}{d x} = y^{3} - x^{3}

Solve Bernoulli ODE:

y = 33 x 3 - ln (x) + 9

y = 33 x 3 - ln (x) + 9

\int \frac{x}{2 x - 1} d x

y^{^{'}} = \frac{x}{10 y}

\int sec (x) tan^{5} (x) d x

\frac{\partial}{\partial x} (x + x y + y)

\frac{2 x}{x ^{2} + 1}, \frac{x}{13}, [- 5, 5]

What is the solution for xy^2(dy)/(dx)=y^3-x^3,y(1)=3 ?
The solution for xy^2(dy)/(dx)=y^3-x^3,y(1)=3 is y=\sqrt[3]{3}x\sqrt[3]{-ln(x)+9}