What is the solution for x^'=x^2+x+1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^'=x^2+x+1

x^{^{'}} = x^{2} + x + 1

x = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

Solution steps

x^{'} (t) = x^{2} + x + 1

Solve separable ODE:

x = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

x = \frac{tan ( \frac{3 ( t + c _{1} )}{2} ) 3}{2} - \frac{1}{2}

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