What is the solution for 8y^{''}+12y^'+5y=0 ?

The solution for 8y^{''}+12y^'+5y=0 is y=e^{-(3t)/4}(c_{1}cos(t/4)+c_{2}sin(t/4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8y^{''}+12y^'+5y=0

8 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

Solution steps

8 y^{''} (t) + 12 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{t}{4}) + c_{2} sin (\frac{t}{4}))

6 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - y = 0

y^{^{''}} + 9 y = 0, y (0) = sin (3 x)

2 y^{^{''}} + y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - 3 y^{^{''}} + 4 y^{^{'}} - 12 y = 0

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

What is the solution for 8y^{''}+12y^'+5y=0 ?
The solution for 8y^{''}+12y^'+5y=0 is y=e^{-(3t)/4}(c_{1}cos(t/4)+c_{2}sin(t/4))