What is the solution for y^{'''}-3y^'=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{'''} (t) - 3 y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{''}} + 10 y^{^{'}} = - 21 y

2 (D^{4} + 11 D^{3} - 4 D^{2} - 69 D + 34) y = 0

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

4 y^{^{''}} + 32 y^{^{'}} + 113 y = 0

y^{^{''}} + 18 y^{^{'}} + 80 y = 0, y (0) = 8, y^{^{'}} (0) = - 76

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