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

The solution for y^'=b(1-y)y is y=-(e^{bt+c_{1}})/(1-e^{bt+c_{1)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=b(1-y)y

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

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

Solution steps

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

Solve separable ODE:

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

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

247 x - 19 x^{2}

\int {y (2 a x + b y)} d x

x \to - 3 lim (\frac{2}{x + 3})

\frac{d}{d x} (\frac{x ^{2} - 2 x + 5}{x ^{2} + 2 x - 3})

\int \frac{4 x ^{5} - 4}{x ^{2} - 4} d x

What is the solution for y^'=b(1-y)y ?
The solution for y^'=b(1-y)y is y=-(e^{bt+c_{1}})/(1-e^{bt+c_{1)}}