What is the solution for 16y^{''}+40y^'+50y=0 ?

The solution for 16y^{''}+40y^'+50y=0 is y=e^{-(5t)/4}(c_{1}cos((5t)/4)+c_{2}sin((5t)/4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

16y^{''}+40y^'+50y=0

16 y^{^{''}} + 40 y^{^{'}} + 50 y = 0

y = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4}))

Solution steps

16 y^{''} (t) + 40 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4}))

y = e^{- \frac{5 t}{4}} (c_{1} cos (\frac{5 t}{4}) + c_{2} sin (\frac{5 t}{4}))

y^{^{''}} - 16 y = 0, y (0) = e^{4 x}, y (0) = 9, y^{^{'}} (0) = 3

y^{^{''}} + y^{^{'}} + 24 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

0.001 y^{^{''}} + 10 y = 0

y^{^{''}} + 1 y - 2 y^{^{'}} = 0

1 y^{^{''}} + 3 y^{^{'}} - 4 y = 0

What is the solution for 16y^{''}+40y^'+50y=0 ?
The solution for 16y^{''}+40y^'+50y=0 is y=e^{-(5t)/4}(c_{1}cos((5t)/4)+c_{2}sin((5t)/4))