What is the solution for y^{''}+4y^'+32y=0 ?

The solution for y^{''}+4y^'+32y=0 is y=e^{-2t}(c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t))

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y^{''}+4y^'+32y=0

y^{^{''}} + 4 y^{^{'}} + 32 y = 0

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

y = e^{- 2 t} (c_{1} cos (27 t) + c_{2} sin (27 t))

y^{^{''}} + 169 y = 0

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = - 2

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

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

y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 28, y^{^{'}} (0) = 2

What is the solution for y^{''}+4y^'+32y=0 ?
The solution for y^{''}+4y^'+32y=0 is y=e^{-2t}(c_{1}cos(2sqrt(7)t)+c_{2}sin(2sqrt(7)t))