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

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

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y^'=y-y^3

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

sin (3 t) cos (5 t)

(2 x - 1) d x + (5 y + 8) d y = 0

\int_{0}^{\frac{1}{3}} \frac{3}{1 + 9 x ^{2}} d x

\int_{- 5}^{1} ∣ x + 3 ∣ d x

\frac{\partial ^{2}}{\partial y ^{2}} (x^{2} + y^{2})

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