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}

y^{^{'}} - y = 6 t e^{2 t}

t an g e n t f (x) = 1 - 4 x, at x = - 1

(2 x y^{2} + 7 y) + (2 x^{2} y + 7 x) y^{^{'}} = 0

x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 = 0

\frac{d}{d x} (y^{2} - x^{2})

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}