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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y^'+4y=t

y^{^{''}} + 4 y^{^{'}} + 4 y = t

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t}{4} - \frac{1}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = t

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t}{4} - \frac{1}{4}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{t}{4} - \frac{1}{4}

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