What is the solution for 2y^{''}+12y^'+68y=0 ?

The solution for 2y^{''}+12y^'+68y=0 is y=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))

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2y^{''}+12y^'+68y=0

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

2 y^{''} (t) + 12 y^{'} (t) + 68 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y^{^{''}} + 14 y = 0

34 y + 10 y^{^{'}} + y^{^{''}} = 0, y (0) = - 3, y^{^{'}} (0) = 6

y^{^{''}} - 25 y = 0, y (0) = e^{5 x}

y^{^{''}} + 24 y = 0

6 y^{^{''}} - y^{^{'}} - 15 y = 0

What is the solution for 2y^{''}+12y^'+68y=0 ?
The solution for 2y^{''}+12y^'+68y=0 is y=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))