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

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

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y^'=2y-y^3

y^{^{'}} = 2 y - y^{3}

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int \frac{1}{x ^{2} - 3} d x

y^{^{'}} = x + 3 y

(1 - 2 x)^{\frac{1}{2}}

\int \frac{1}{( x + 1 ) x ^{2} + 2 x - 3} d x

\int (\frac{1 + cos ( 2 x )}{2})^{2} d x

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