What is the solution for y^{''}+2y^'+y=5e^{-t} ?

The solution for y^{''}+2y^'+y=5e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(5e^{-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=5e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 5 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5 e ^{- t} t ^{2}}{2}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 5 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5 e ^{- t} t ^{2}}{2}

y = c_{1} e^{- t} + c_{2} t e^{- t} + \frac{5 e ^{- t} t ^{2}}{2}

y^{^{''}} + y = - cos (2 t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 50 x sin (x) + 100 x cos (x)

y^{^{''}} - 4 y = - 7 e^{2 x} + x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 3 y = e^{2 x} sin (x)

y^{^{''}} + 9 y = e^{2 x}

What is the solution for y^{''}+2y^'+y=5e^{-t} ?
The solution for y^{''}+2y^'+y=5e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+(5e^{-t}t^2)/2