What is the solution for 13y+6y^'+y^{''}=0 ?

The solution for 13y+6y^'+y^{''}=0 is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))

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13y+6y^'+y^{''}=0

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = - 3

3 x^{^{''}} + 75 x = 0, x (0) = 0, x^{^{'}} (0) = - 10

12 y^{^{''}} - 13 y^{^{'}} + y = 0, y (0) = 12, y^{^{'}} (0) = \frac{1}{12}

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

What is the solution for 13y+6y^'+y^{''}=0 ?
The solution for 13y+6y^'+y^{''}=0 is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))