What is the solution for x^{''}+2x^'+4x=0 ?

The solution for x^{''}+2x^'+4x=0 is x=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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x^{''}+2x^'+4x=0

x^{^{''}} + 2 x^{^{'}} + 4 x = 0

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 4 x = 0

Solve linear ODE:

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

x = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\frac{d}{d z} (\frac{z + i}{z - i})

x \to - 1 lim (3 x^{2} - 2 x^{2} + 4)

e^{3} e^{t} (- sin (π t))

y^{^{'}} = y^{2} - 4

\int_{1}^{R} x^{- \frac{8}{5}} d x

What is the solution for x^{''}+2x^'+4x=0 ?
The solution for x^{''}+2x^'+4x=0 is x=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))