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

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

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y^'=2y(1-y)

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{\partial}{\partial x} (\frac{( 2 x + 2 y )}{- 6 x - 8 y})

f (x) = 8 arcsin (x^{4})

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\frac{d}{d x} (3 x^{2} - 18 x + 15)

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