What is the solution for y^{''}+4y^'+40y=0 ?

The solution for y^{''}+4y^'+40y=0 is y=e^{-2t}(c_{1}cos(6t)+c_{2}sin(6t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+40y=0

y^{^{''}} + 4 y^{^{'}} + 40 y = 0

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 40 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{- 2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

What is the solution for y^{''}+4y^'+40y=0 ?
The solution for y^{''}+4y^'+40y=0 is y=e^{-2t}(c_{1}cos(6t)+c_{2}sin(6t))