What is the solution for 16y^{''}-8y^'+5y=0 ?

The solution for 16y^{''}-8y^'+5y=0 is y=e^{t/4}(c_{1}cos(t/2)+c_{2}sin(t/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16y^{''}-8y^'+5y=0

16 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

y = e^{\frac{t}{4}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

Solution steps

16 y^{''} (t) - 8 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{\frac{t}{4}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

y = e^{\frac{t}{4}} (c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2}))

4 y^{^{''}} + 36 y = 0

36 y^{^{'''}} + 60 y^{^{''}} + 25 y^{^{'}} = 0

h^{^{''}} - 2 h^{^{'}} + 10 h = 0, h (0) = 1, h^{^{'}} (0) = 10

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

y^{^{''}} + y = 0, y (\frac{π}{2}) = 0, y (0) = 1

What is the solution for 16y^{''}-8y^'+5y=0 ?
The solution for 16y^{''}-8y^'+5y=0 is y=e^{t/4}(c_{1}cos(t/2)+c_{2}sin(t/2))