What is the solution for ([y-y^2])/t =y^',y(1)=2 ?

The solution for ([y-y^2])/t =y^',y(1)=2 is y=-(2t)/(1-2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

([y-y^2])/t =y^',y(1)=2

\frac{[ y - y ^{2} ]}{t} = y^{^{'}}, y (1) = 2

y = - \frac{2 t}{1 - 2 t}

Solution steps

\frac{[ y - y ^{2} ]}{t} = y^{'} (t)

Solve separable ODE:

y = - \frac{2 t}{1 - 2 t}

y = - \frac{2 t}{1 - 2 t}

x \to 0 lim (x (11 + \frac{4}{x}))

\int cos (\frac{nπ}{2} x) d x

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} - 2 x})

x \to 1 lim (sin (\frac{1}{x}))

\int (x^{2} \cdot sin (x)) d x

What is the solution for ([y-y^2])/t =y^',y(1)=2 ?
The solution for ([y-y^2])/t =y^',y(1)=2 is y=-(2t)/(1-2t)