What is the solution for y'=y^3-y ?

The solution for y'=y^3-y is y=sqrt(1/(-e^{2t+c_{1))+1}},y=-sqrt(1/(-e^{2t+c_{1))+1}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y'=y^3-y

y^{^{'}} = y^{3} - y

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

Solution steps

y^{'} (t) = y^{3} - y

Solve Bernoulli ODE:

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

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

\frac{d}{d x} (3 x^{4} + \frac{1}{3 x ^{4}} + \frac{3}{2 x ^{2}})

\frac{d}{d x} (arcsin (\frac{3 x}{4}))

\frac{d x}{d t} = \frac{5}{x e ^{t + 5 x}}

\int_{- 5}^{- 2} 4 x^{3} d x

\frac{\partial}{\partial x} (5 x^{2} y^{2})

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