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

The solution for x^{''}+2x^'+2x=2 is x=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

Solution steps

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

Solve linear ODE:

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

x = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + 1

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