What is the solution for y^2y^'=2(t+ty^3)e^{t^2},y(0)=0 ?

The solution for y^2y^'=2(t+ty^3)e^{t^2},y(0)=0 is y=\sqrt[3]{e^{3e^{t^2}-3}-1}

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y^2y^'=2(t+ty^3)e^{t^2},y(0)=0

y^{2} y^{^{'}} = 2 (t + t y^{3}) e^{t^{2}}, y (0) = 0

Solution steps

y^{2} y^{'} (t) = 2 (t + t y^{3}) e^{t^{2}}

Solve separable ODE:

y^{^{'}} (x) = \frac{2 x - 1}{3 y ^{2} - 3}, y (1) = 0

\frac{d y}{d x} = \frac{2}{y}

\frac{d y}{d x} = \frac{( x ^{2} + 2 )}{y}

y^{^{'}} = e^{- y} sin (x), y (\frac{π}{2}) = \frac{1}{2}

y^{^{'}} + 2 y = 0, y (0) = 1.5

What is the solution for y^2y^'=2(t+ty^3)e^{t^2},y(0)=0 ?
The solution for y^2y^'=2(t+ty^3)e^{t^2},y(0)=0 is y=\sqrt[3]{e^{3e^{t^2}-3}-1}