What is the solution for x^{''}+2x^'+3x=0 ?

The solution for x^{''}+2x^'+3x=0 is x=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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

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

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

Solution steps

x^{''} (t) + 2 x^{'} (t) + 3 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^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 2, y^{^{'}} (0) = 10

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} = 7 y^{^{'}} - 12 y

(D^{2} + 4) y = 0

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

What is the solution for x^{''}+2x^'+3x=0 ?
The solution for x^{''}+2x^'+3x=0 is x=e^{-t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))