What is the solution for 3y^{''}-2y^'+6y=0 ?

The solution for 3y^{''}-2y^'+6y=0 is y=e^{t/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

3y^{''}-2y^'+6y=0

3 y^{^{''}} - 2 y^{^{'}} + 6 y = 0

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

Solution steps

3 y^{''} (t) - 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

y = e^{\frac{t}{3}} (c_{1} cos (\frac{17 t}{3}) + c_{2} sin (\frac{17 t}{3}))

What is the solution for 3y^{''}-2y^'+6y=0 ?
The solution for 3y^{''}-2y^'+6y=0 is y=e^{t/3}(c_{1}cos((sqrt(17)t)/3)+c_{2}sin((sqrt(17)t)/3))