What is the solution for 9y^{''}+6y^'+82y=0,y(0)=2 ?

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

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9y^{''}+6y^'+82y=0,y(0)=2

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = 2

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

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

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

(4 D^{2} - 3 D - 10) y = 0

x^{^{'''}} - 4 x^{^{''}} - 3 x^{^{'}} + 18 x = 0

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

y^{^{''}} - 10 y^{^{'}} - y = 0

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