en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y(y-1)

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

tangent of f(x)=(4x+3)/(x+1),\at x=3 tangent of

f (x) = \frac{4 x + 3}{x + 1}, at x = 3

derivative of (arcsec((x^2/2))^{3x})

\frac{d}{d x} ((arcsec (\frac{x ^{2}}{2}))^{3 x})

f (x) = - \frac{x}{2}

derivative of x+5 derivative of

x + 5

area y=3x,y=4x^2 area

y = 3 x, y = 4 x^{2}

Frequently Asked Questions (FAQ)

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