What is the solution for 4y^{''}+20y^'+29y=0,y(0)=8 ?

The solution for 4y^{''}+20y^'+29y=0,y(0)=8 is y=e^{-(5t)/2}(8cos(t)+c_{2}sin(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}+20y^'+29y=0,y(0)=8

4 y^{^{''}} + 20 y^{^{'}} + 29 y = 0, y (0) = 8

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

y = e^{- \frac{5 t}{2}} (8 cos (t) + c_{2} sin (t))

What is the solution for 4y^{''}+20y^'+29y=0,y(0)=8 ?
The solution for 4y^{''}+20y^'+29y=0,y(0)=8 is y=e^{-(5t)/2}(8cos(t)+c_{2}sin(t))