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

The solution for 2y^{''}+2y^'+8y=0 is y=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 3, y^{^{'}} (0) = 5

y^{^{''}} - y = 0, y^{^{'}} (0) = 0

y = \frac{d ^{2} y}{d x ^{2}}

What is the solution for 2y^{''}+2y^'+8y=0 ?
The solution for 2y^{''}+2y^'+8y=0 is y=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))