What is the solution for x^{''}+2x^'+2x=2,x(0)=0,x^'(0)=0 ?

The solution for x^{''}+2x^'+2x=2,x(0)=0,x^'(0)=0 is x=e^{-t}(-cos(t)-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(0)=0,x^'(0)=0

x^{^{''}} + 2 x^{^{'}} + 2 x = 2, x (0) = 0, x^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

What is the solution for x^{''}+2x^'+2x=2,x(0)=0,x^'(0)=0 ?
The solution for x^{''}+2x^'+2x=2,x(0)=0,x^'(0)=0 is x=e^{-t}(-cos(t)-sin(t))+1