What is the solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 ?

The solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{-t}sin(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

y = - \frac{7}{2} e^{- t} sin (2 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = - \frac{7}{2} e^{- t} sin (2 t)

y = - \frac{7}{2} e^{- t} sin (2 t)

What is the solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 ?
The solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{-t}sin(2t)