What is the solution for y^'=ky*(100-y) ?

The solution for y^'=ky*(100-y) is y=(100)/(e^{-100kt+c_{1)}+1}

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y^'=ky*(100-y)

y^{^{'}} = k y \cdot (100 - y)

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

Solution steps

y^{'} (t) = k y \cdot (100 - y)

Solve Bernoulli ODE:

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

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What is the solution for y^'=ky*(100-y) ?
The solution for y^'=ky*(100-y) is y=(100)/(e^{-100kt+c_{1)}+1}