What is the solution for x^{''}+4x^'+16x=0 ?

The solution for x^{''}+4x^'+16x=0 is x=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+4x^'+16x=0

x^{^{''}} + 4 x^{^{'}} + 16 x = 0

x = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

x^{''} (t) + 4 x^{'} (t) + 16 x = 0

Solve linear ODE:

x = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

x = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

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

y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0

y^{^{''}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\frac{3 y ^{^{''}} + 2 y ^{^{'}} + y}{4} = 0

What is the solution for x^{''}+4x^'+16x=0 ?
The solution for x^{''}+4x^'+16x=0 is x=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))