What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1 ?

The solution for 4y^{''}+20y^'+25y=0,y(0)=1 is y=e^{-(5t)/2}+c_{2}e^{-(5t)/2}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}+20y^'+25y=0,y(0)=1

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1

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

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

What is the solution for 4y^{''}+20y^'+25y=0,y(0)=1 ?
The solution for 4y^{''}+20y^'+25y=0,y(0)=1 is y=e^{-(5t)/2}+c_{2}e^{-(5t)/2}t