What is the solution for y^{''}+2y^'+2y=0,y(0)=0,y^'(0)=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+2y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{- t} sin (t)

Solution steps

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

Solve linear ODE:

y = e^{- t} sin (t)

y = e^{- t} sin (t)

\frac{\partial}{\partial y} (\frac{x + y}{y + z})

\int 2 sin (2 x) d x

\int (1 - x) x d x

e^{- \frac{x}{2}} e^{- \frac{1}{4}}

\frac{\partial}{\partial x} (6 y cos (x))

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