What is the solution for y^{''}+2y^'+63y=0 ?

The solution for y^{''}+2y^'+63y=0 is y=e^{-t}(c_{1}cos(sqrt(62)t)+c_{2}sin(sqrt(62)t))

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y^{''}+2y^'+63y=0

y^{^{''}} + 2 y^{^{'}} + 63 y = 0

y = e^{- t} (c_{1} cos (62 t) + c_{2} sin (62 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 63 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (62 t) + c_{2} sin (62 t))

y = e^{- t} (c_{1} cos (62 t) + c_{2} sin (62 t))

4 x^{^{''}} + 12 x^{^{'}} + 9 x = 0

y^{^{'''}} + 10 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 7 y^{^{'}} + 10 y = 0, y^{^{'}} (0) = - \frac{1}{3}

(D^{2} - 6 D + 25) y = 0

What is the solution for y^{''}+2y^'+63y=0 ?
The solution for y^{''}+2y^'+63y=0 is y=e^{-t}(c_{1}cos(sqrt(62)t)+c_{2}sin(sqrt(62)t))