en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=ky(100-y)

Solution

y^{^{'}} = k y (100 - y)

Solution

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

Solution steps

y^{'} (t) = k y (t) (100 - y)

Solve Bernoulli ODE:

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

y = \frac{100}{e ^{- 100 k t + c_{1}} + 1}

Popular Examples

derivative of 11sin(x)

\frac{d}{d x} (11 sin (x))

(\partial)/(\partial x)(2y+z^3-xz)

\frac{\partial}{\partial x} (2 y + z^{3} - x z)

derivative of y=(2x)/(x+1)derivative of

y = \frac{2 x}{x + 1}

7(sin(t)y^'+cos(t)y)=cos(t)sin^3(t)

7 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{3} (t)

(dy}{dx}=\frac{-2x)/y ,y(0)=1

\frac{d y}{d x} = \frac{- 2 x}{y}, y (0) = 1

Frequently Asked Questions (FAQ)

What is the solution for y^'=ky(100-y) ?
The solution for y^'=ky(100-y) is y=(100)/(e^{-100kt+c_{1)}+1}