What is the solution for y^{''}+4y^'+6.25y=0 ?

The solution for y^{''}+4y^'+6.25y=0 is y=e^{-2t}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

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y^{''}+4y^'+6.25y=0

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 6.25 y = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 3 y^{^{'}} + y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 9

y^{^{''}} - 196 y = 0

25 y^{^{''}} - 60 y^{^{'}} + 36 y = 0

y^{^{'''}} + 7 y^{^{''}} + 19 y^{^{'}} + 13 y = 0

What is the solution for y^{''}+4y^'+6.25y=0 ?
The solution for y^{''}+4y^'+6.25y=0 is y=e^{-2t}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))