What is the solution for y'=0.002y(2500-y) ?

The solution for y'=0.002y(2500-y) is y= 1/(-0.2e^{-5t-c_{1)}+0.0004}

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y'=0.002y(2500-y)

y^{^{'}} = 0.002 y (2500 - y)

y = \frac{1}{- 0.2 e ^{- 5 t - c_{1}} + 0.0004}

Solution steps

y^{'} (t) = 0.002 y (2500 - y)

Solve Bernoulli ODE:

y = \frac{1}{- 0.2 e ^{- 5 t - c_{1}} + 0.0004}

y = \frac{1}{- 0.2 e ^{- 5 t - c_{1}} + 0.0004}

\frac{d ^{2}}{d x ^{2}} (x^{3} + y^{3})

π^{2}

\int \frac{2}{3 x - 2} d x

\frac{d}{d x} ((sec (2 x))^{7})

5 - x^{2}, 2 - 2 x

What is the solution for y'=0.002y(2500-y) ?
The solution for y'=0.002y(2500-y) is y= 1/(-0.2e^{-5t-c_{1)}+0.0004}