What is the solution for 2x^{''}+4x^'+8x=0 ?

The solution for 2x^{''}+4x^'+8x=0 is x=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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2x^{''}+4x^'+8x=0

2 x^{^{''}} + 4 x^{^{'}} + 8 x = 0

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

Solution steps

2 x^{''} (t) + 4 x^{'} (t) + 8 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))

What is the solution for 2x^{''}+4x^'+8x=0 ?
The solution for 2x^{''}+4x^'+8x=0 is x=e^{-t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))