What is the solution for y^'-2y=-4y^3 ?

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

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y^'-2y=-4y^3

y^{^{'}} - 2 y = - 4 y^{3}

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int x^{3} e^{2} d x

\int (1 + t)^{2} d t

\frac{2}{x + 5}, at x = 4

f (x) = 2 x^{2} - x - 3, at x = 1

y = 2 - 6 x

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