What is the solution for 2y^{''}+12y^'+19y=0 ?

The solution for 2y^{''}+12y^'+19y=0 is y=e^{-3t}(c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2))))

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2y^{''}+12y^'+19y=0

2 y^{^{''}} + 12 y^{^{'}} + 19 y = 0

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

Solution steps

2 y^{''} (t) + 12 y^{'} (t) + 19 y = 0

Solve linear ODE:

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

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

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

60 x^{^{''}} + 240 x = 0

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

y^{^{''}} + (\frac{1}{2}) y = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = e^{5 x}

What is the solution for 2y^{''}+12y^'+19y=0 ?
The solution for 2y^{''}+12y^'+19y=0 is y=e^{-3t}(c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2))))