What is the solution for y^{''}+6y^'+8y=0 ?

The solution for y^{''}+6y^'+8y=0 is y=c_{1}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

Popular >

y^{''}+6y^'+8y=0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0

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

Solution steps

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

Solve linear ODE:

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

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

\int x^{- 4} + x^{\frac{1}{3}} d x

s (t) = 3 t + 1

\frac{\partial}{\partial y} (x^{2} yz - x y z^{3})

\frac{d}{d y} (\frac{y}{x})

\frac{d}{d x} (\frac{2 x ^{7} + 5 x ^{3} - 1}{4})

What is the solution for y^{''}+6y^'+8y=0 ?
The solution for y^{''}+6y^'+8y=0 is y=c_{1}e^{-2t}+c_{2}e^{-4t}