What is the solution for x^{''}+2x^'+6x=0 ?

The solution for x^{''}+2x^'+6x=0 is x=e^{-t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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x^{''}+2x^'+6x=0

x^{^{''}} + 2 x^{^{'}} + 6 x = 0

x = e^{- t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 6 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (5 t) + c_{2} sin (5 t))

x = e^{- t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y^{^{''''}} + 20 y^{^{''}} - 100 y = 0

y^{^{''}} - 3 y^{^{'}} + 7 y = 0

y^{^{'''}} - 5 y^{^{''}} + 17 y - 13 y = 0

\frac{d ^{2} f}{d x ^{2}} = a f (x)

y^{^{'''}} + 36 y^{^{'}} = 0, y^{^{''}} (0) = 108

What is the solution for x^{''}+2x^'+6x=0 ?
The solution for x^{''}+2x^'+6x=0 is x=e^{-t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))