What is the solution for x^{''}+2*x^'+10x=0 ?

The solution for x^{''}+2*x^'+10x=0 is x=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+2*x^'+10x=0

x^{^{''}} + 2 \cdot x^{^{'}} + 10 x = 0

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

x^{''} (t) + 2 \cdot x^{'} (t) + 10 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y^{^{''}} - 121 y = 0

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

y^{^{''}} - 14 y^{^{'}} + 45 y = 0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = 5

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0

What is the solution for x^{''}+2*x^'+10x=0 ?
The solution for x^{''}+2*x^'+10x=0 is x=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))