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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{x + 2}{x - 1}

\frac{d y}{d x} = \frac{cos ( 5 x )}{e ^{5 y}}

\int \frac{2 e ^{2 x}}{1 + e ^{2 x}} d x

y \frac{d y}{d x} = 2 x (64 + y^{2})

\frac{d}{d x} (- \frac{3 x}{y})

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