en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y(2+y)t

Solution

y^{^{'}} = y (2 + y) t

Solution

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

Solution steps

y^{'} (t) = y (2 + y) t

Solve Bernoulli ODE:

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

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

Graph

Popular Examples

derivative of sin^x(x)

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

integral of 1/(xsqrt(x^2+1))

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

derivative of ln(2x-4)

\frac{d}{d x} (ln (2 x - 4))

integral from 0 to 1 of (1/(1+x^2))

\int_{0}^{1} (\frac{1}{1 + x ^{2}}) d x

limit as x approaches 6 of 4

x \to 6 lim (4)

Frequently Asked Questions (FAQ)

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