What is the solution for y^{'''}+2y^{''}-8y^'=0 ?

The solution for y^{'''}+2y^{''}-8y^'=0 is y=c_{1}+c_{2}e^{2t}+c_{3}e^{-4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{'''} (t) + 2 y^{''} (t) - 8 y^{'} (t) = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 0, y^{^{'}} (0) = 0

13 y^{^{''}} - 14 y^{^{'}} + y = 0, y (0) = 13, y^{^{'}} (0) = \frac{1}{13}

y^{^{''}} - \frac{1}{4} y = 0

y^{^{''}} + a^{2} \cdot n^{2} \cdot π^{2} \cdot y = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d x ^{2}} + k^{2} \cdot y^{^{'}} = 0

What is the solution for y^{'''}+2y^{''}-8y^'=0 ?
The solution for y^{'''}+2y^{''}-8y^'=0 is y=c_{1}+c_{2}e^{2t}+c_{3}e^{-4t}