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

The solution for y^{''}+y^'+5y=0 is y=e^{-t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

\int_{- 1}^{1} (x - 1) d x

\int - 8 x cos (3 x) d x

\int e^{- θ} cos (9 θ) d θ

(4 x^{2}) e^{- 10 x}

x \to \frac{π}{2} lim (\frac{tan ( x ) - sec ( x )}{cos ( x )})

What is the solution for y^{''}+y^'+5y=0 ?
The solution for y^{''}+y^'+5y=0 is y=e^{-t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))