What is the solution for y^'=0.0016y(2500-y) ?

The solution for y^'=0.0016y(2500-y) is y= 4/(-e^{-4t-c_{1)}+0.0016}

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y^'=0.0016y(2500-y)

y^{^{'}} = 0.0016 y (2500 - y)

y = \frac{4}{- e ^{- 4 t - c_{1}} + 0.0016}

Solution steps

y^{'} (t) = 0.0016 y (2500 - y)

Solve Bernoulli ODE:

y = \frac{4}{- e ^{- 4 t - c_{1}} + 0.0016}

y = \frac{4}{- e ^{- 4 t - c_{1}} + 0.0016}

\int \frac{1}{x ^{2} + 2 x + 122} d x

\int_{0}^{\infty} \frac{7}{x ^{2} + 4} d x

\frac{\partial}{\partial x} (y^{4} cos (4 x) \cdot 4)

\int \frac{x ^{2} - 49}{x} d x

x + y^{2} = 2, x + y = 0

What is the solution for y^'=0.0016y(2500-y) ?
The solution for y^'=0.0016y(2500-y) is y= 4/(-e^{-4t-c_{1)}+0.0016}