What is the solution for 9y^{''}+6y^'+82y=0 ?

The solution for 9y^{''}+6y^'+82y=0 is y=e^{-t/3}(c_{1}cos(3t)+c_{2}sin(3t))

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9y^{''}+6y^'+82y=0

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0

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

Solution steps

9 y^{''} (t) + 6 y^{'} (t) + 82 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 7 y^{^{'}} - 6 y = 0

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

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

2 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

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

What is the solution for 9y^{''}+6y^'+82y=0 ?
The solution for 9y^{''}+6y^'+82y=0 is y=e^{-t/3}(c_{1}cos(3t)+c_{2}sin(3t))