What is the solution for y^{''}+2y^'+y=12e^{-t} ?

The solution for y^{''}+2y^'+y=12e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+6t^2e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=12e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 12 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 6 t^{2} e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 12 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + 6 t^{2} e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 6 t^{2} e^{- t}

\frac{\partial}{\partial x} (x e^{x} + e^{x})

\frac{\partial}{\partial y} (x - 2 y)

y^{^{''}} - 14 y^{^{'}} + 50 y = 0, y (0) = 10, y (\frac{π}{2}) = 3

f (x) = \frac{e ^{x} x - e ^{x}}{x ^{2}}

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

What is the solution for y^{''}+2y^'+y=12e^{-t} ?
The solution for y^{''}+2y^'+y=12e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+6t^2e^{-t}