What is the solution for y^{''}+14y^'+53y=0 ?

The solution for y^{''}+14y^'+53y=0 is y=e^{-7t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}+14y^'+53y=0

y^{^{''}} + 14 y^{^{'}} + 53 y = 0

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) + 14 y^{'} (t) + 53 y = 0

Solve linear ODE:

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{- 7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

4 y^{^{''}} - 4 y^{^{'}} - 5 y = 0

(d^{3} - 3 d^{2} - 10 d) y = 0

0.2 y^{^{''}} + 14 y^{^{'}} + \frac{1}{0.004} y = 0

(D^{2} + 2 D + 5)^{3} y = 0

66 y^{^{'''}} + 17 y^{^{''}} + y^{^{'}} = 0

What is the solution for y^{''}+14y^'+53y=0 ?
The solution for y^{''}+14y^'+53y=0 is y=e^{-7t}(c_{1}cos(2t)+c_{2}sin(2t))