What is the solution for y^{''}+2y^'+2y=5e^t+2t+2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+2y=5e^t+2t+2

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 e^{t} + 2 t + 2

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} = 3 y^{^{''}} + 9 e^{3} x + 5 cos (x)

y^{^{''}} + 6 y^{^{'}} + 9 y = 3 sin (x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 9 y = 10 e^{-} 1, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 24 e^{x}

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} sin (x)

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