What is the solution for y^{''}+4y^'+4y=4 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=4

y^{^{''}} + 4 y^{^{'}} + 4 y = 4

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = e^{3} x

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

y^{^{''}} - 9 y = - 8 e^{t}

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 x} sin (4 x)

y^{^{'''}} + 3 y^{^{''}} + y^{^{'}} + 3 y = e^{x} + 4

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