What is the solution for x*y^2*(dy)/(dx)=y^3-4*x^3,y(1)=2 ?

The solution for x*y^2*(dy)/(dx)=y^3-4*x^3,y(1)=2 is y=\sqrt[3]{4}x\sqrt[3]{-3ln(x)+2}

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xy^2(dy)/(dx)=y^3-4*x^3,y(1)=2

x \cdot y^{2} \cdot \frac{d y}{d x} = y^{3} - 4 \cdot x^{3}, y (1) = 2

y = 34 x 3 - 3 ln (x) + 2

Solution steps

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

Solve Bernoulli ODE:

y = 34 x 3 - 3 ln (x) + 2

y = 34 x 3 - 3 ln (x) + 2

x^{2} - 8, 7 x

x \to 0 lim (\frac{sin ( 5 x )}{5})

y^{^{''}} - 2 y^{^{'}} - 8 y = x e^{x} + 4

x \to \frac{π}{2} - lim (cos (x))

\frac{d y}{d x} = \frac{( 5 y ^{3} - 2 x y )}{x ^{2}}

What is the solution for x*y^2*(dy)/(dx)=y^3-4*x^3,y(1)=2 ?
The solution for x*y^2*(dy)/(dx)=y^3-4*x^3,y(1)=2 is y=\sqrt[3]{4}x\sqrt[3]{-3ln(x)+2}