What is the solution for y^{''}+6y^'+13y=10e^{-2t} ?

The solution for y^{''}+6y^'+13y=10e^{-2t} is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))+2e^{-2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'+13y=10e^{-2t}

y^{^{''}} + 6 y^{^{'}} + 13 y = 10 e^{- 2 t}

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

Solution steps

y^{''} (t) + 6 y^{'} (t) + 13 y = 10 e^{- 2 t}

Solve linear ODE:

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

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

y^{^{''}} + y = t + t sin (t)

y^{^{''}} + 11 y^{^{'}} + 18 y = 5 sin (3 x)

y^{^{''}} - 9 y^{^{'}} = 81 x^{2} + 7

y^{^{''}} + y = sec^{3} (x), y (0) = 1

- y^{^{''}} + 10 y^{^{'}} + 25 y = 30 x + 3

What is the solution for y^{''}+6y^'+13y=10e^{-2t} ?
The solution for y^{''}+6y^'+13y=10e^{-2t} is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))+2e^{-2t}