What is the solution for x^{''}+2x^'+x=e^{-2t} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+2x^'+x=e^{-2t}

x^{^{''}} + 2 x^{^{'}} + x = e^{- 2 t}

x = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

Solution steps

x^{''} (t) + 2 x^{'} (t) + x = e^{- 2 t}

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

x = c_{1} e^{- t} + c_{2} t e^{- t} + e^{- 2 t}

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (5 t)

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

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

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

y^{^{''}} - 4 y = 3 x e^{2 x}

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