What is the solution for y^{''}+16y^'+73y=0,y(0)=4 ?

The solution for y^{''}+16y^'+73y=0,y(0)=4 is y=e^{-8t}(4cos(3t)+c_{2}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+16y^'+73y=0,y(0)=4

y^{^{''}} + 16 y^{^{'}} + 73 y = 0, y (0) = 4

y = e^{- 8 t} (4 cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 16 y^{'} (t) + 73 y = 0

Solve linear ODE:

y = e^{- 8 t} (4 cos (3 t) + c_{2} sin (3 t))

y = e^{- 8 t} (4 cos (3 t) + c_{2} sin (3 t))

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 2

y^{^{''}} + \frac{y}{4} = 0

y^{^{''}} - y^{^{'}} + 144 y = 0

y^{^{''}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+16y^'+73y=0,y(0)=4 ?
The solution for y^{''}+16y^'+73y=0,y(0)=4 is y=e^{-8t}(4cos(3t)+c_{2}sin(3t))