What is the solution for y^{''}+8y^'+20y=0 ?

The solution for y^{''}+8y^'+20y=0 is y=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+20y=0

y^{^{''}} + 8 y^{^{'}} + 20 y = 0

y = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{- 4 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

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

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

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3, y^{^{'}} (0) = - 12

3 y^{^{'''}} - 12 y^{^{'}} - 4 y^{^{''}} + y^{^{''''}} = 0

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

What is the solution for y^{''}+8y^'+20y=0 ?
The solution for y^{''}+8y^'+20y=0 is y=e^{-4t}(c_{1}cos(2t)+c_{2}sin(2t))