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

The solution for x^{''}+x^'+5x=0 is x=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 >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

a x^{^{''}} + 2 x^{^{'}} = 0

y^{^{''}} + \frac{y ^{^{'}}}{2} - \frac{y}{2} = 0

\frac{d ^{3} y}{d x ^{3}} - 8 y = 0

y^{^{''''}} - 24 y^{^{'''}} + 144 y^{^{''}} = 0

(D^{4} - D^{3} - 7 D^{2} + 3 D) y = 0

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