What is the solution for y^{''}+5y^'+20y=0 ?

The solution for y^{''}+5y^'+20y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(55)t)/2)+c_{2}sin((sqrt(55)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'+20y=0

y^{^{''}} + 5 y^{^{'}} + 20 y = 0

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{55 t}{2}) + c_{2} sin (\frac{55 t}{2}))

Solution steps

y^{''} (t) + 5 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{55 t}{2}) + c_{2} sin (\frac{55 t}{2}))

y = e^{- \frac{5 t}{2}} (c_{1} cos (\frac{55 t}{2}) + c_{2} sin (\frac{55 t}{2}))

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 4, y^{^{'}} (0) = - 6

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 3 y + 2 y = 0

y^{^{''}} + 9 y = 0, y^{^{'}} (\frac{π}{2}) = 0, y (0) = 0

y^{^{''}} + y^{^{'}} - 6 y = 0, y (1) = 2 e^{2} - e^{- 3}

What is the solution for y^{''}+5y^'+20y=0 ?
The solution for y^{''}+5y^'+20y=0 is y=e^{-(5t)/2}(c_{1}cos((sqrt(55)t)/2)+c_{2}sin((sqrt(55)t)/2))