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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

Solution steps

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

Solve linear ODE:

y = c_{1} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

y = c_{1} + e^{- t} (c_{2} cos (t) + c_{3} sin (t))

arctan (\frac{y}{x})

(sin (t))^{^{'}}

f (x) = 4 x^{2} - 15

e^{2 x} y^{^{'}} - 4 y = 0

\frac{d}{d x} (x^{\frac{8}{9}})

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