What is the solution for x^{''}+6x^'+8x=0,x(0)=7 ?

The solution for x^{''}+6x^'+8x=0,x(0)=7 is x=(7-c_{2})e^{-2t}+c_{2}e^{-4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^{''}+6x^'+8x=0,x(0)=7

x^{^{''}} + 6 x^{^{'}} + 8 x = 0, x (0) = 7

x = (7 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

Solution steps

x^{''} (t) + 6 x^{'} (t) + 8 x = 0

Solve linear ODE:

x = (7 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

x = (7 - c_{2}) e^{- 2 t} + c_{2} e^{- 4 t}

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

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

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

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

y^{^{''}} + 0.25 y = 0

What is the solution for x^{''}+6x^'+8x=0,x(0)=7 ?
The solution for x^{''}+6x^'+8x=0,x(0)=7 is x=(7-c_{2})e^{-2t}+c_{2}e^{-4t}