What is the solution for y^{''}+10y^'+5y=0 ?

The solution for y^{''}+10y^'+5y=0 is y=c_{1}e^{(-5+2sqrt(5))t}+c_{2}e^{(-5-2sqrt(5))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+10y^'+5y=0

y^{^{''}} + 10 y^{^{'}} + 5 y = 0

y = c_{1} e^{(- 5 + 25) t} + c_{2} e^{(- 5 - 25) t}

Solution steps

y^{''} (t) + 10 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = c_{1} e^{(- 5 + 25) t} + c_{2} e^{(- 5 - 25) t}

y = c_{1} e^{(- 5 + 25) t} + c_{2} e^{(- 5 - 25) t}

y^{^{''}} + y^{^{'}} + 12 y = 0, y (0) = 1

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

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 2 y^{^{'}} + 4 y = 0, y (0) = 1, y (0) = - 2

y^{^{''}} - 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 1

What is the solution for y^{''}+10y^'+5y=0 ?
The solution for y^{''}+10y^'+5y=0 is y=c_{1}e^{(-5+2sqrt(5))t}+c_{2}e^{(-5-2sqrt(5))t}