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

The solution for linear y^{'''}-2y^{''}-y^'+2y=0 is y=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

Popular >

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

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 20 y^{^{'}} + 116 y = 0

9 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 2

2 y^{^{''}} + 12 y^{^{'}} + 19 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 32 y^{^{'}} + 160 y = 0

4 q^{^{''}} + 8 q^{^{'}} + \frac{1}{0.0125} q = 88 q

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