What is the solution for y^'=0.001y(400-y) ?

The solution for y^'=0.001y(400-y) is y=(400}{-1000e^{\frac{-c_{1}-2t)/5}+1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=0.001y(400-y)

y^{^{'}} = 0.001 y (400 - y)

y = \frac{400}{- 1000 e ^{\frac{- c _{1} - 2 t}{5}} + 1}

Solution steps

y^{'} (t) = 0.001 y (400 - y)

Solve Bernoulli ODE:

y = \frac{400}{- 1000 e ^{\frac{- c _{1} - 2 t}{5}} + 1}

y = \frac{400}{- 1000 e ^{\frac{- c _{1} - 2 t}{5}} + 1}

\int 25 sec^{- 2} (x t a) n^{3} x d x

\int 9 sec^{6} (x) d x

n = 0 \sum \infty \frac{3}{7}

\int \frac{1}{2} cos (\frac{1}{2}) t d t

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

What is the solution for y^'=0.001y(400-y) ?
The solution for y^'=0.001y(400-y) is y=(400}{-1000e^{\frac{-c_{1}-2t)/5}+1}