What is the solution for 4y^{''}+8y^'+5y=0 ?

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

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4y^{''}+8y^'+5y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} - 8 y^{^{'}} - 33 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

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

3 y^{^{'''}} + 10 y^{^{''}} + 15 y^{^{'}} + 4 y = 0

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