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))

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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}))

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

3 y^{^{''}} + 11 y^{^{'}} - 4 y = 0, y (0) = 4, y^{^{'}} (0) = - \frac{5}{3}

y^{^{''}} + 20 y = 0

(D^{2} + 4 D + 5) y = 0

y^{^{''}} + 2 y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 6

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))