What is the solution for y^{''}+2y^'+9y=0 ?

The solution for y^{''}+2y^'+9y=0 is y=e^{-t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+9y=0

y^{^{''}} + 2 y^{^{'}} + 9 y = 0

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y = e^{- t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y^{^{''}} - 2 y^{^{'}} - 2 y = 0, y (0) = 0, y^{^{'}} (0) = 3

2 y^{^{''}} - 3 y + y = 0

y^{^{''}} - cy = 0

y^{^{''}} - 13 y^{^{'}} + 40 y = 0

4 y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 3

What is the solution for y^{''}+2y^'+9y=0 ?
The solution for y^{''}+2y^'+9y=0 is y=e^{-t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))