What is the solution for y^{'''}+y^{''}+4y^'+4y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{'''} (t) + y^{''} (t) + 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

x^{^{''}} + a x^{^{'}} + b x = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 2

\frac{d ^{2} y}{d t ^{2}} + 6 \frac{d y}{d t} + 9 y = 0

2 \frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - y = 0

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

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