What is the solution for x^{''}+2x^'+5x=0 ?

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

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x^{''}+2x^'+5x=0

x^{^{''}} + 2 x^{^{'}} + 5 x = 0

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 5 x = 0

Solve linear ODE:

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

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

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 6, y (1) = 8

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 0

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 0

2 y^{^{''}} + 3 y^{^{'}} + y = 0

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