What is the solution for y^'=(t(4y-y^2))/3 ?

The solution for y^'=(t(4y-y^2))/3 is y=-(4e^{(2t^2)/3+4c_{1}})/(1-e^{(2t^2)/3+4c_{1)}}

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y^'=(t(4y-y^2))/3

y^{^{'}} = \frac{t ( 4 y - y ^{2} )}{3}

y = - \frac{4 e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}{1 - e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}

Solution steps

y^{'} (t) = \frac{t ( 4 y - y ^{2} )}{3}

Solve separable ODE:

y = - \frac{4 e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}{1 - e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}

y = - \frac{4 e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}{1 - e ^{\frac{2 t ^{2}}{3} + 4 c_{1}}}

What is the solution for y^'=(t(4y-y^2))/3 ?
The solution for y^'=(t(4y-y^2))/3 is y=-(4e^{(2t^2)/3+4c_{1}})/(1-e^{(2t^2)/3+4c_{1)}}