What is the solution for 2y^{''}+2y^'+5y=0 ?

The solution for 2y^{''}+2y^'+5y=0 is y=e^{-t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+2y^'+5y=0

2 y^{^{''}} + 2 y^{^{'}} + 5 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

2 y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

16 \frac{d ^{4} y}{d x ^{4}} - y = 0

x^{^{''}} + 8 x^{^{'}} + 4 x = 0, x (0) = 0, x^{^{'}} (0) = 1

x^{^{''}} - 4 a^{2} x = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 10

y^{^{''''}} + 3 y^{^{'}} = 0

What is the solution for 2y^{''}+2y^'+5y=0 ?
The solution for 2y^{''}+2y^'+5y=0 is y=e^{-t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))