What is the solution for y^{''}+4y^'+4y=3e^{-t} ?

The solution for y^{''}+4y^'+4y=3e^{-t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+3e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=3e^{-t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + 3 e^{- t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 3 e^{- t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + 3 e^{- t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + 3 e^{- t}

y^{^{''}} - 4 y + 8 y = 4 sin (x)

y^{^{''}} - 8 y^{^{'}} \cdot + 25 = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 75 t^{3}

y^{^{''}} - 7 y^{^{'}} + 12 y = 5 sin (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 18 + e^{x}

What is the solution for y^{''}+4y^'+4y=3e^{-t} ?
The solution for y^{''}+4y^'+4y=3e^{-t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}+3e^{-t}