en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y(1-y),y(0)=-2

Solution

y^{^{'}} = y (1 - y), y (0) = - 2

Solution

y = \frac{2 e ^{t}}{- 3 + 2 e ^{t}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 e ^{t}}{- 3 + 2 e ^{t}}

y = \frac{2 e ^{t}}{- 3 + 2 e ^{t}}

Popular Examples

integral of (y+1)/(y^2+y+1)

\int \frac{y + 1}{y ^{2} + y + 1} d y

(\partial)/(\partial z)(xcos(y)sin(z))

\frac{\partial}{\partial z} (x cos (y) sin (z))

integral of (x^3)/(sqrt(x-1))

\int \frac{x ^{3}}{x - 1} d x

limit as x approaches 2 of (x+6)/(x-2)

x \to 2 lim (\frac{x + 6}{x - 2})

integral of 3x^2sqrt((2x^3+5))

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=y(1-y),y(0)=-2 ?
The solution for y^'=y(1-y),y(0)=-2 is y=(2e^t)/(-3+2e^t)