What is the solution for 9y^{''}+18y^'+13y=0 ?

The solution for 9y^{''}+18y^'+13y=0 is y=e^{-t}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))

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9y^{''}+18y^'+13y=0

9 y^{^{''}} + 18 y^{^{'}} + 13 y = 0

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

Solution steps

9 y^{''} (t) + 18 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

q^{^{''}} + 20 q^{^{'}} + 100 q = 0

\frac{d ^{2} y}{d x ^{2}} - 366 \frac{d y}{d x} = 0

16 y^{^{''}} - 9 y = 0

5 y^{^{'''}} + 45 y^{^{'}} - 1350 y = 0

\frac{d ^{2} y}{d t ^{2}} + 4 \cdot y = 0

What is the solution for 9y^{''}+18y^'+13y=0 ?
The solution for 9y^{''}+18y^'+13y=0 is y=e^{-t}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))