What is the solution for 3y^{''}+12y^'+39y=0 ?

The solution for 3y^{''}+12y^'+39y=0 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))

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3y^{''}+12y^'+39y=0

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 0

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

Solution steps

3 y^{''} (t) + 12 y^{'} (t) + 39 y = 0

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = 0, y (0) = cos (x)

3 V^{^{''}} + 4 V^{^{'}} + 3 V = 0

x^{^{''}} + 4 a^{2} x = 0

q^{^{''}} + 806 q^{^{'}} + 8000 q = 0

y^{^{'''}} - 12 y^{^{''}} + 21 y^{^{'}} + 98 y = 0

What is the solution for 3y^{''}+12y^'+39y=0 ?
The solution for 3y^{''}+12y^'+39y=0 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))