What is the solution for 8y^{''}+26y=0 ?

The solution for 8y^{''}+26y=0 is y=c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

8y^{''}+26y=0

8 y^{^{''}} + 26 y = 0

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

Solution steps

8 y^{''} (t) + 26 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

y = c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2})

What is the solution for 8y^{''}+26y=0 ?
The solution for 8y^{''}+26y=0 is y=c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2)