What is the solution for 3y^{''}+12y^'+24y=0 ?

The solution for 3y^{''}+12y^'+24y=0 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))

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3y^{''}+12y^'+24y=0

3 y^{^{''}} + 12 y^{^{'}} + 24 y = 0

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

Solution steps

3 y^{''} (t) + 12 y^{'} (t) + 24 y = 0

Solve linear ODE:

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

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

- f^{^{''}} (t) - c f (t) = 0

6 y^{^{''}} - 7 y^{^{'}} + y = 0, y (0) = 6, y^{^{'}} (0) = \frac{1}{6}

y^{^{''}} + 10 y^{^{'}} + 21 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} + 9.25 y = 0, y^{^{'}} (0) = 8

x^{^{''}} + 9 x = 0, x (0) = 1

What is the solution for 3y^{''}+12y^'+24y=0 ?
The solution for 3y^{''}+12y^'+24y=0 is y=e^{-2t}(c_{1}cos(2t)+c_{2}sin(2t))