What is the solution for y^{''}+4y^'+4y=e^{-t},y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y^'+4y=e^{-t},y(0)=1,y^'(0)=0 is y=e^{-2t}t+e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=e^{-t},y(0)=1,y^'(0)=0

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- t}, y (0) = 1, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = e^{- t}

Solve linear ODE:

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

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

y^{^{''}} + 9 y = - 12 sec (3 t)

y^{^{''}} + 4 y = g (t), y (0) = 5, y^{^{'}} (0) = 16

x^{^{''}} - x = t

4 y^{^{''}} - 4 y^{^{'}} + y = 16 e^{\frac{1}{2} t}

16 y^{^{''''}} - y = e^{\frac{x}{2}}

What is the solution for y^{''}+4y^'+4y=e^{-t},y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y^'+4y=e^{-t},y(0)=1,y^'(0)=0 is y=e^{-2t}t+e^{-t}