What is the solution for y^'=(2y-1)^3 ?

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

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y^'=(2y-1)^3

y^{^{'}} = (2 y - 1)^{3}

y = \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}, y = - \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}

Solution steps

y^{'} (t) = (2 y - 1)^{3}

Solve separable ODE:

y = \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}, y = - \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}

y = \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}, y = - \frac{- \frac{1}{4 ( t + c _{1} )}}{2} + \frac{1}{2}

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