What is the solution for y^{''}+2y^'+5y=5 ?

The solution for y^{''}+2y^'+5y=5 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+5y=5

y^{^{''}} + 2 y^{^{'}} + 5 y = 5

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 5

Solve linear ODE:

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

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

y^{^{''}} + 9 y = cos (9 x)

y^{^{''}} - y^{^{'}} - 6 y = 32 cos (3 x)

y^{^{'''}} - 3 y^{^{'}} + 2 y = e^{- 2 x}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 3

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

What is the solution for y^{''}+2y^'+5y=5 ?
The solution for y^{''}+2y^'+5y=5 is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+1