What is the solution for x^{''}+2ax^'+bx=0 ?

The solution for x^{''}+2ax^'+bx=0 is x=c_{1}e^{(-a+sqrt(a^2-b))t}+c_{2}e^{(-a-sqrt(a^2-b))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+2ax^'+bx=0

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

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

Solution steps

x^{''} (t) + 2 a x^{'} (t) + b x (t) = 0

Solve linear ODE:

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

x = c_{1} e^{(- a + a^{2} - b) t} + c_{2} e^{(- a - a^{2} - b) t}

y^{^{''''}} + 4 y^{^{'''}} + 4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

x^{^{''}} - 6 x^{^{''}} + x = 0

y^{^{'''}} + 9 y^{^{'}} = 0, y (0) = 5, y^{^{'}} (0) = 4, y^{^{''}} (0) = 3

y^{^{''}} - \frac{2}{5} y = 0

y^{^{''''}} + 2 y^{^{'''}} - 4 y^{^{''}} - 2 y^{^{'}} + \frac{1}{2} y = 0

What is the solution for x^{''}+2ax^'+bx=0 ?
The solution for x^{''}+2ax^'+bx=0 is x=c_{1}e^{(-a+sqrt(a^2-b))t}+c_{2}e^{(-a-sqrt(a^2-b))t}