What is the solution for y^{''}+8y^'+32y=0,y^'0=-4,y^0=1 ?

The solution for y^{''}+8y^'+32y=0,y^'0=-4,y^0=1 is y=e^{-4t}(c_{1}cos(4t)+c_{2}sin(4t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+32y=0,y^'0=-4,y^0=1

y^{^{''}} + 8 y^{^{'}} + 32 y = 0, y^{^{'}} 0 = - 4, y^{0} (0) = 1

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = - 1, y^{^{'}} (0) = 8

4 y^{^{''}} + 3 y^{^{''}} + y = 0

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

y^{^{''}} - 5 y^{^{'}} - 14 y = 0, y (0) = 0, y^{^{'}} (0) = 3

4 y^{^{''}} - 4 y^{^{'}} + (- 3) y = 0, y (0) = 5, y^{^{'}} (0) = 0

What is the solution for y^{''}+8y^'+32y=0,y^'0=-4,y^0=1 ?
The solution for y^{''}+8y^'+32y=0,y^'0=-4,y^0=1 is y=e^{-4t}(c_{1}cos(4t)+c_{2}sin(4t))