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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial y} (\frac{x ^{3}}{x ^{2} + y ^{2}})

n = 1 \sum \infty (- 5 x)^{n}

\int (x + 4) (2 x + 1) d x

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

\frac{d}{d x} (cos (θ) x)

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