What is the solution for y^{''}+20y^'+500y=0 ?

The solution for y^{''}+20y^'+500y=0 is y=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+20y^'+500y=0

y^{^{''}} + 20 y^{^{'}} + 500 y = 0

y = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

Solution steps

y^{''} (t) + 20 y^{'} (t) + 500 y = 0

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

y = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t))

x d y + y d x = 0

\int π t - t^{2} cos (n t) d t

\frac{\frac{2}{3}}{s ^{2} + 2 s + 17}

\frac{25 - x ^{2}}{32 x}

\int (tan (5 x) - sec (5 x))^{2} d x

What is the solution for y^{''}+20y^'+500y=0 ?
The solution for y^{''}+20y^'+500y=0 is y=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))