What is the solution for y^{''}+2y^'+y=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=1

y^{^{''}} + 2 y^{^{'}} + y = 1

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 1

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} + 9 y = 3 sin (3 t)

x \to + 2 lim (\frac{1}{2})

\frac{d}{d t} (t^{2} t)

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 4) e^{x})

ln (1 + 9 x)

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