What is the solution for y^'=0.0012y(1900-y) ?

The solution for y^'=0.0012y(1900-y) is y=(2.28)/(-e^{-2.27999…t-c_{1)}+0.0012}

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y^'=0.0012y(1900-y)

y^{^{'}} = 0.0012 y (1900 - y)

y = \frac{2.28}{- e ^{- 2.27999 \dots t - c_{1}} + 0.0012}

Solution steps

y^{'} (t) = 0.0012 y (1900 - y)

Solve Bernoulli ODE:

y = \frac{2.28}{- e ^{- 2.27999 \dots t - c_{1}} + 0.0012}

y = \frac{2.28}{- e ^{- 2.27999 \dots t - c_{1}} + 0.0012}

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\frac{d y}{d x}, y^{\frac{2}{3}} - x^{2} + x y = 1

\frac{\partial}{\partial x} (5 x sin (3 x^{2} y))

\frac{d}{d x} (sin^{l n (x)} (x))

x \to 2 lim (5 x + 1)

What is the solution for y^'=0.0012y(1900-y) ?
The solution for y^'=0.0012y(1900-y) is y=(2.28)/(-e^{-2.27999…t-c_{1)}+0.0012}