What is the solution for 4y^{''}-y^'+17y=0 ?

The solution for 4y^{''}-y^'+17y=0 is y=e^{t/8}(c_{1}cos((sqrt(271)t)/8)+c_{2}sin((sqrt(271)t)/8))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}-y^'+17y=0

4 y^{^{''}} - y^{^{'}} + 17 y = 0

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

Solution steps

4 y^{''} (t) - y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

What is the solution for 4y^{''}-y^'+17y=0 ?
The solution for 4y^{''}-y^'+17y=0 is y=e^{t/8}(c_{1}cos((sqrt(271)t)/8)+c_{2}sin((sqrt(271)t)/8))