What is the solution for y^{''}-20y^'+12y=0 ?

The solution for y^{''}-20y^'+12y=0 is y=c_{1}e^{2(5+sqrt(22))t}+c_{2}e^{2(5-sqrt(22))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-20y^'+12y=0

y^{^{''}} - 20 y^{^{'}} + 12 y = 0

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

Solution steps

y^{''} (t) - 20 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

y = c_{1} e^{2 (5 + 22) t} + c_{2} e^{2 (5 - 22) t}

What is the solution for y^{''}-20y^'+12y=0 ?
The solution for y^{''}-20y^'+12y=0 is y=c_{1}e^{2(5+sqrt(22))t}+c_{2}e^{2(5-sqrt(22))t}