What is the solution for 1/20 y^{''}+3y^'+80y=0 ?

The solution for 1/20 y^{''}+3y^'+80y=0 is y=e^{-30t}(c_{1}cos(10sqrt(7)t)+c_{2}sin(10sqrt(7)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/20 y^{''}+3y^'+80y=0

\frac{1}{20} y^{^{''}} + 3 y^{^{'}} + 80 y = 0

y = e^{- 30 t} (c_{1} cos (107 t) + c_{2} sin (107 t))

Solution steps

\frac{1}{20} y^{''} (t) + 3 y^{'} (t) + 80 y = 0

Solve linear ODE:

y = e^{- 30 t} (c_{1} cos (107 t) + c_{2} sin (107 t))

y = e^{- 30 t} (c_{1} cos (107 t) + c_{2} sin (107 t))

y^{^{''}} - 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 3

7 x^{^{''}} + 3 x^{^{''}} + 7 x = 0, x (0) = - 4, x^{^{'}} (0) = 0

y^{^{'''}} + y^{^{''}} - 56 y^{^{'}} = 0

18 y^{^{''}} + (45 - 27) y^{^{'}} - 57 y = 0

y^{^{''}} - 10 y + 24 y = 0

What is the solution for 1/20 y^{''}+3y^'+80y=0 ?
The solution for 1/20 y^{''}+3y^'+80y=0 is y=e^{-30t}(c_{1}cos(10sqrt(7)t)+c_{2}sin(10sqrt(7)t))