What is the solution for y^{''}+3y^'+6y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+3y^'+6y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(15))e^{-(3t)/2}sin((sqrt(15)t)/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+3y^'+6y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 3 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

Solution steps

y^{''} (t) + 3 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

y = \frac{2}{15} e^{- \frac{3 t}{2}} sin (\frac{15 t}{2})

What is the solution for y^{''}+3y^'+6y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'+6y=0,y(0)=0,y^'(0)=1 is y= 2/(sqrt(15))e^{-(3t)/2}sin((sqrt(15)t)/2)