What is the solution for x^{'''}-2x^{''}-x^'+2x=0 ?

The solution for x^{'''}-2x^{''}-x^'+2x=0 is x=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^{'''}-2x^{''}-x^'+2x=0

x^{^{'''}} - 2 x^{^{''}} - x^{^{'}} + 2 x = 0

x = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

Solution steps

x^{'''} (t) - 2 x^{''} (t) - x^{'} (t) + 2 x = 0

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

x = c_{1} e^{2 t} + c_{2} e^{- t} + c_{3} e^{t}

What is the solution for x^{'''}-2x^{''}-x^'+2x=0 ?
The solution for x^{'''}-2x^{''}-x^'+2x=0 is x=c_{1}e^{2t}+c_{2}e^{-t}+c_{3}e^t