What is the solution for 8y^{''}-2y^'-y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

8y^{''}-2y^'-y=0

8 y^{^{''}} - 2 y^{^{'}} - y = 0

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

Solution steps

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

Solve linear ODE:

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

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

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