What is the solution for y^{''}+2y^'+3y=0,y(0)=6,y^'(0)=-2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+3y=0,y(0)=6,y^'(0)=-2

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y = e^{- t} (6 cos (2 t) + 22 sin (2 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = e^{- t} (6 cos (2 t) + 22 sin (2 t))

y = e^{- t} (6 cos (2 t) + 22 sin (2 t))

u^{^{''}} + a^{2} H (t) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.459

25 x^{^{''}} + 10 x^{^{'}} + 226 x = 0

- 5 y^{^{'}} + 2 y^{^{''}} - y^{^{'}} - 6 y = 0

121 y^{^{''}} - 154 y^{^{'}} + 74 y = 0

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