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

The solution for y^{''''}+2y^{'''}-y^{''}-2y^'=0 is y=c_{1}+c_{2}e^{-2t}+c_{3}e^{-t}+c_{4}e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

Solution steps

y^{(4)} (t) + 2 y^{'''} (t) - y^{''} (t) - 2 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

y = c_{1} + c_{2} e^{- 2 t} + c_{3} e^{- t} + c_{4} e^{t}

What is the solution for y^{''''}+2y^{'''}-y^{''}-2y^'=0 ?
The solution for y^{''''}+2y^{'''}-y^{''}-2y^'=0 is y=c_{1}+c_{2}e^{-2t}+c_{3}e^{-t}+c_{4}e^t