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Popular >

y'=y(2-y)

Solution

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

Solution

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

Solution steps

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

Solve Bernoulli ODE:

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

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

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Frequently Asked Questions (FAQ)

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