What is the solution for y^'=y(1000-y) ?

The solution for y^'=y(1000-y) is y=(1000)/(-e^{-1000t-c_{1)}+1}

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y^'=y(1000-y)

y^{^{'}} = y (1000 - y)

y = \frac{1000}{- e ^{- 1000 t - c_{1}} + 1}

Solution steps

y^{'} (t) = y (1000 - y)

Solve Bernoulli ODE:

y = \frac{1000}{- e ^{- 1000 t - c_{1}} + 1}

y = \frac{1000}{- e ^{- 1000 t - c_{1}} + 1}

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What is the solution for y^'=y(1000-y) ?
The solution for y^'=y(1000-y) is y=(1000)/(-e^{-1000t-c_{1)}+1}