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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 0 lim (x \cdot e^{x})

\frac{d}{d x} (\frac{10 lo g _{4} ( x )}{x})

x \to \infty lim (\frac{6 ^{x + 1}}{7 ^{x - 1}})

\int (x - 3) sin (π x) d x

\int \frac{cot ( x )}{x} d x

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