What is the solution for y^'=0.0015y(2700-y) ?

The solution for y^'=0.0015y(2700-y) is y=(4.05)/(-e^{-4.05t-c_{1)}+0.0015}

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y^'=0.0015y(2700-y)

y^{^{'}} = 0.0015 y (2700 - y)

y = \frac{4.05}{- e ^{- 4.05 t - c_{1}} + 0.0015}

Solution steps

y^{'} (t) = 0.0015 y (2700 - y)

Solve Bernoulli ODE:

y = \frac{4.05}{- e ^{- 4.05 t - c_{1}} + 0.0015}

y = \frac{4.05}{- e ^{- 4.05 t - c_{1}} + 0.0015}

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\frac{\partial}{\partial x} (1 - 2 x^{2} - 2 y)

What is the solution for y^'=0.0015y(2700-y) ?
The solution for y^'=0.0015y(2700-y) is y=(4.05)/(-e^{-4.05t-c_{1)}+0.0015}