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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}

Graph

Popular Examples

integral of (-28t+40)

\int (- 28 t + 40) d t

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x \to 1 - lim (1 - x^{2})

3sin(x)sin(y)+cos(y)y^'=0

3 sin (x) sin (y) + cos (y) y^{^{'}} = 0

integral of 4x-3

\int 4 x - 3 d x

integral of sin(8x)cos(5x)

\int sin (8 x) cos (5 x) d x

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}