What is the solution for 36y^{''}+108y^'+97y=0 ?

The solution for 36y^{''}+108y^'+97y=0 is y=e^{-(3t)/2}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))

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36y^{''}+108y^'+97y=0

36 y^{^{''}} + 108 y^{^{'}} + 97 y = 0

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

Solution steps

36 y^{''} (t) + 108 y^{'} (t) + 97 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 2 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = a, y^{^{'}} (0) = - 15

2 y^{^{''}} + 128 y = 0

y^{^{''''}} - y^{^{'''}} - y^{^{''}} - y^{^{'}} - 2 y = 0

What is the solution for 36y^{''}+108y^'+97y=0 ?
The solution for 36y^{''}+108y^'+97y=0 is y=e^{-(3t)/2}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))