What is the solution for y^{''}+4y^'+13y=e^{-t} ?

The solution for y^{''}+4y^'+13y=e^{-t} is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+1/10 e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+13y=e^{-t}

y^{^{''}} + 4 y^{^{'}} + 13 y = e^{- t}

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{10} e^{- t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = e^{- t}

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{10} e^{- t}

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{1}{10} e^{- t}

y^{^{''}} + 5 y^{^{'}} - 6 y = 2 e^{- x}

y^{^{''}} + 3 y^{^{'}} - y = t

y^{^{''}} - 5 y^{^{'}} + 4 y = 3 e^{x} sin (x)

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} (12 x^{2} - 6 x)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} = x - 8 e^{x}

What is the solution for y^{''}+4y^'+13y=e^{-t} ?
The solution for y^{''}+4y^'+13y=e^{-t} is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+1/10 e^{-t}