What is the solution for y^2y^'=y^3-3x,y(0)=2 ?

The solution for y^2y^'=y^3-3x,y(0)=2 is y=\sqrt[3]{3x+1+7e^{3x}}

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y^2y^'=y^3-3x,y(0)=2

y^{2} y^{^{'}} = y^{3} - 3 x, y (0) = 2

y = 3 3 x + 1 + 7 e^{3 x}

Solution steps

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

Solve Bernoulli ODE:

y = 3 3 x + 1 + 7 e^{3 x}

y = 3 3 x + 1 + 7 e^{3 x}

x \to 0 lim (2 (1 + 3 x)^{\frac{2}{x}})

x \to 8 lim (x^{\frac{4}{3}})

\frac{d}{d x} (\frac{3 x ^{2} + 2 x}{4 x - 2})

ln (3 + x)

\int \frac{1}{x ^{2} x ^{2} - 25} d x

What is the solution for y^2y^'=y^3-3x,y(0)=2 ?
The solution for y^2y^'=y^3-3x,y(0)=2 is y=\sqrt[3]{3x+1+7e^{3x}}