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

The solution for y^'=sqrt(y)(1-sqrt(y)) is y=e^{-t+c_{1}}+2e^{(-t+c_{1})/2}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=sqrt(y)(1-sqrt(y))

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

y = e^{- t + c_{1}} + 2 e^{\frac{- t + c _{1}}{2}} + 1

Solution steps

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

Solve for

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = y (1 - y)

Isolate

t^{^{'}} : t^{^{'}} = - \frac{1}{y ( y - 1 )}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int - \frac{1}{y ( y - 1 )} d y

\int - \frac{1}{y ( y - 1 )} d y = - 2 ln (y - 1) + c_{1}

t = - 2 ln (y - 1) + c_{1}

Isolate

y : y = e^{- t + c_{1}} + 2 e^{\frac{- t + c _{1}}{2}} + 1

y = e^{- t + c_{1}} + 2 e^{\frac{- t + c _{1}}{2}} + 1

What is the solution for y^'=sqrt(y)(1-sqrt(y)) ?
The solution for y^'=sqrt(y)(1-sqrt(y)) is y=e^{-t+c_{1}}+2e^{(-t+c_{1})/2}+1