What is the solution for 9y^{''}+4y^'+5y=0 ?

The solution for 9y^{''}+4y^'+5y=0 is y=e^{-(2t)/9}(c_{1}cos((sqrt(41)t)/9)+c_{2}sin((sqrt(41)t)/9))

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9y^{''}+4y^'+5y=0

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

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

y = e^{- \frac{2 t}{9}} (c_{1} cos (\frac{41 t}{9}) + c_{2} sin (\frac{41 t}{9}))

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - 3

y^{^{''''}} = y

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

y = y^{^{''}}

y^{^{''}} + 6 y^{^{'}} + 3 y = 0

What is the solution for 9y^{''}+4y^'+5y=0 ?
The solution for 9y^{''}+4y^'+5y=0 is y=e^{-(2t)/9}(c_{1}cos((sqrt(41)t)/9)+c_{2}sin((sqrt(41)t)/9))