What is the solution for y^{'''}-8y^{''}-y^'+8y=0 ?

The solution for y^{'''}-8y^{''}-y^'+8y=0 is y=c_{1}e^{8t}+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 >

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

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

y = c_{1} e^{8 t} + c_{2} e^{- t} + c_{3} e^{t}

Solution steps

y^{'''} (t) - 8 y^{''} (t) - y^{'} (t) + 8 y = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- t} + c_{3} e^{t}

y = c_{1} e^{8 t} + c_{2} e^{- t} + c_{3} e^{t}

a \frac{d ^{2} y}{d x ^{2}} + y b^{2} = 0

y^{^{''}} + \frac{48}{2} y = 0

y^{^{''}} - y^{^{'}} + 0.25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

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

2 y^{^{''}} + 10 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 0, y (0) = 5

What is the solution for y^{'''}-8y^{''}-y^'+8y=0 ?
The solution for y^{'''}-8y^{''}-y^'+8y=0 is y=c_{1}e^{8t}+c_{2}e^{-t}+c_{3}e^t