What is the solution for 9y^{''}+12y^'+29y=0 ?

The solution for 9y^{''}+12y^'+29y=0 is y=e^{-(2t)/3}(c_{1}cos((5t)/3)+c_{2}sin((5t)/3))

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9y^{''}+12y^'+29y=0

9 y^{^{''}} + 12 y^{^{'}} + 29 y = 0

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

Solution steps

9 y^{''} (t) + 12 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

y = e^{- \frac{2 t}{3}} (c_{1} cos (\frac{5 t}{3}) + c_{2} sin (\frac{5 t}{3}))

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - \frac{9}{4}

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 y = 0

What is the solution for 9y^{''}+12y^'+29y=0 ?
The solution for 9y^{''}+12y^'+29y=0 is y=e^{-(2t)/3}(c_{1}cos((5t)/3)+c_{2}sin((5t)/3))