What is the solution for y'=0.0018y(2400-y) ?

The solution for y'=0.0018y(2400-y) is y=(4.32)/(-e^{-4.31999…t-c_{1)}+0.0018}

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y'=0.0018y(2400-y)

y^{^{'}} = 0.0018 y (2400 - y)

y = \frac{4.32}{- e ^{- 4.31999 \dots t - c_{1}} + 0.0018}

Solution steps

y^{'} (t) = 0.0018 y (2400 - y)

Solve Bernoulli ODE:

y = \frac{4.32}{- e ^{- 4.31999 \dots t - c_{1}} + 0.0018}

y = \frac{4.32}{- e ^{- 4.31999 \dots t - c_{1}} + 0.0018}

\frac{\partial}{\partial x} (x^{3} y z^{2})

ln ∣ ln (x) ∣

\frac{d y}{d t} + \frac{1}{20} y = \frac{3}{2}, y (0) = 0

\frac{d}{d t} (t cos (t) - sin (t))

(t^{3})^{^{'}}

What is the solution for y'=0.0018y(2400-y) ?
The solution for y'=0.0018y(2400-y) is y=(4.32)/(-e^{-4.31999…t-c_{1)}+0.0018}