What is the solution for 9y^{''}+6y^'+82y=0,y(0)=-1,y^'(0)=2 ?

The solution for 9y^{''}+6y^'+82y=0,y(0)=-1,y^'(0)=2 is y=e^{-t/3}(-cos(3t)+5/9 sin(3t))

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9y^{''}+6y^'+82y=0,y(0)=-1,y^'(0)=2

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y = e^{- \frac{t}{3}} (- cos (3 t) + \frac{5}{9} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{3}} (- cos (3 t) + \frac{5}{9} sin (3 t))

y = e^{- \frac{t}{3}} (- cos (3 t) + \frac{5}{9} sin (3 t))

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y^{^{''}} - 4 y^{^{'}} - 14 y = 0

6 y^{^{''}} + 5 y^{^{'}} + y = 0

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

(D^{4} + 8 D^{2} + 16) y = 0

What is the solution for 9y^{''}+6y^'+82y=0,y(0)=-1,y^'(0)=2 ?
The solution for 9y^{''}+6y^'+82y=0,y(0)=-1,y^'(0)=2 is y=e^{-t/3}(-cos(3t)+5/9 sin(3t))