What is the solution for 4y^{''}+16y^'+17y=0 ?

The solution for 4y^{''}+16y^'+17y=0 is y=e^{-2t}(c_{1}cos(t/2)+c_{2}sin(t/2))

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4y^{''}+16y^'+17y=0

4 y^{^{''}} + 16 y^{^{'}} + 17 y = 0

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

Solution steps

4 y^{''} (t) + 16 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = - 4, y^{^{'}} (0) = 3

(D^{3} + 2 D^{2} - 15 D) y = 0

y^{^{''}} + b y^{^{'}} + 64 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 10 y^{^{'}} + α y = 0

What is the solution for 4y^{''}+16y^'+17y=0 ?
The solution for 4y^{''}+16y^'+17y=0 is y=e^{-2t}(c_{1}cos(t/2)+c_{2}sin(t/2))