The title says it all. How do I find the undamped position function in general for 2nd order ODE with a forcing function equal to $0$? Here is an example. I am working on a problem $$\frac{1}{2}x'' + 3x'+4x=0$$
$$ \to x''+6x'+8x=0$$
the damped position function is $x(t)= 4e^{-2t}-2e^{-4t}$
I have no idea how the undamped position function becomes $2\cos(2\sqrt{2}t)$
If I got a complex root for an answer I think I get partially there. But still wont get the answer entirely. Since this problem has no complex roots I don't even know where to start to get the undamped position function. When I have a complex number for roots it looks like all I do is take what ever I have for $cos$ as the answer and subtract $\mathbb{\alpha}$ from x, ($cos(ax-\mathbb{\alpha})$ but I know that is not right. I need clarification.
Best Answer
The question asks for the undamped position. Thus the (undamped) equations of motion are
$$\frac{1}{2} x^{\prime \prime} (t) + 4 x(t) = 0 $$
which has the solution $x(t) = \cos{(2\sqrt{2}t)} + c$.