What is the solution for y^'=y(1-0.0002y) ?

The solution for y^'=y(1-0.0002y) is y=-(e^{t+c_{1}})/(1-0.0002e^{t+c_{1)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y(1-0.0002y)

y^{^{'}} = y (1 - 0.0002 y)

y = - \frac{e ^{t + c_{1}}}{1 - 0.0002 e ^{t + c_{1}}}

Solution steps

y^{'} (t) = y (1 - 0.0002 y)

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{1 - 0.0002 e ^{t + c_{1}}}

y = - \frac{e ^{t + c_{1}}}{1 - 0.0002 e ^{t + c_{1}}}

\frac{1}{( s + 2 ) ( s ^{2} - 2 s + 5 )}

\int_{- 3}^{- 1} x^{2} e^{x^{3} + 27} d x

\int tan^{2} (a x) cos (a x) d x

\int e^{x} arctan (e^{x}) d x

x^{2} + 8 x

What is the solution for y^'=y(1-0.0002y) ?
The solution for y^'=y(1-0.0002y) is y=-(e^{t+c_{1}})/(1-0.0002e^{t+c_{1)}}