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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

y^{^{''''}} + 8 y^{^{'''}} + 16 y^{^{''}} = 0

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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