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))

\frac{d}{d x} (\frac{3}{x ^{2} - 5})

f (x) = 8 x - x^{2}, (1, 7)

k = 0 \sum \infty sin (k)

2^{40}

6 \frac{d y}{d x} = y^{2}

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))