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

The solution for y^'=1+y+y^2 is y=(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

Popular >

y^'=1+y+y^2

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{x ^{3}}{2 y ^{2}}

\frac{d y}{d x} = - \frac{0.1}{cos ( y )}, y (0) = \frac{π}{2}

\frac{d v}{d t} = \frac{t + 2}{v}

\frac{d y}{d x} = \frac{y ( 2 - 3 x )}{x ( 1 - 3 y )}

\frac{d y}{d x} = 8 x y^{2}, y (0) = 1

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