What is the solution for y^{''}+2y^'-8y=0,y(0)=3,y(0)=-12 ?

The solution for y^{''}+2y^'-8y=0,y(0)=3,y(0)=-12 is y=(3-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

y^{''}+2y^'-8y=0,y(0)=3,y(0)=-12

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = 3, y (0) = - 12

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

Solution steps

y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

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

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

What is the solution for y^{''}+2y^'-8y=0,y(0)=3,y(0)=-12 ?
The solution for y^{''}+2y^'-8y=0,y(0)=3,y(0)=-12 is y=(3-c_{2})e^{2t}+c_{2}e^{-4t}