What is the solution for 5y^{''}+20y^'+640y=0 ?

The solution for 5y^{''}+20y^'+640y=0 is y=e^{-2t}(c_{1}cos(2sqrt(31)t)+c_{2}sin(2sqrt(31)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5y^{''}+20y^'+640y=0

5 y^{^{''}} + 20 y^{^{'}} + 640 y = 0

y = e^{- 2 t} (c_{1} cos (231 t) + c_{2} sin (231 t))

Solution steps

5 y^{''} (t) + 20 y^{'} (t) + 640 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (231 t) + c_{2} sin (231 t))

y = e^{- 2 t} (c_{1} cos (231 t) + c_{2} sin (231 t))

y^{^{''}} - 19 y^{^{'}} + 10 y = 0

16 y^{^{''}} + 56 y^{^{'}} + 49 y = 0, y (0) = 5, y^{^{'}} (0) = 5

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

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

x^{^{''}} + \frac{2}{16} x^{^{'}} + 2 x = 0

What is the solution for 5y^{''}+20y^'+640y=0 ?
The solution for 5y^{''}+20y^'+640y=0 is y=e^{-2t}(c_{1}cos(2sqrt(31)t)+c_{2}sin(2sqrt(31)t))