What is the solution for 4y^{''}+9y^'+9y=0 ?

The solution for 4y^{''}+9y^'+9y=0 is y=e^{-(9t)/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))

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4y^{''}+9y^'+9y=0

4 y^{^{''}} + 9 y^{^{'}} + 9 y = 0

y = e^{- \frac{9 t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

Solution steps

4 y^{''} (t) + 9 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{9 t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

y = e^{- \frac{9 t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

y^{^{''}} - 10 y^{^{'}} - 144 y = 0, y (0) = 9, y^{^{'}} (0) = 58

y^{^{''}} + a \cdot y^{^{'}} - b \cdot y = 0

y^{^{''}} - 10 y^{^{'}} + 6 y = 0

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

4 y^{^{''}} + 4 y^{^{'}} - 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

What is the solution for 4y^{''}+9y^'+9y=0 ?
The solution for 4y^{''}+9y^'+9y=0 is y=e^{-(9t)/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))