What is the solution for 1/8 y^{''}+4y^'+12y=0 ?

The solution for 1/8 y^{''}+4y^'+12y=0 is y=c_{1}e^{4(sqrt(10)-4)t}+c_{2}e^{-4(4+sqrt(10))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

1/8 y^{''}+4y^'+12y=0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 12 y = 0

y = c_{1} e^{4 (10 - 4) t} + c_{2} e^{- 4 (4 + 10) t}

Solution steps

\frac{1}{8} y^{''} (t) + 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{4 (10 - 4) t} + c_{2} e^{- 4 (4 + 10) t}

y = c_{1} e^{4 (10 - 4) t} + c_{2} e^{- 4 (4 + 10) t}

What is the solution for 1/8 y^{''}+4y^'+12y=0 ?
The solution for 1/8 y^{''}+4y^'+12y=0 is y=c_{1}e^{4(sqrt(10)-4)t}+c_{2}e^{-4(4+sqrt(10))t}