The roots of this differential equation

Question

In my book it says for the motion of the spring the ODE is $mx''+kx'=0$ and its auxiliary equation is $mr^2+k=0$, thus the roots are $r=\pm\sqrt{\dfrac{k}{m}}i$. Haven't they assumed here that the roots of the auxiliary equation aren't real and distinct? What if they are, for some values of $m$ and $k$. Would the general solution they have given still work which is $x(t) = A\sin(u)t + B\cos(u)t$ .... Because I often encounter questions like this where the values of $m$ and $k$ are not always defined and don't know where to go after finding the auxiliary equation.

Answer 1

$m$ and $k$ are non-negative, because antigravific matter and passive anti-springs do not exist.

Then

$$mr^2+k=0$$ can only have imaginary roots.

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In the general case, for whatever root $z=x+iy$, a solution is

$$e^{zt}=e^{xt}+e^{iyt}=e^{xt}(\cos yt+i\sin yt).$$

As the complex roots always appear in pairs, the final expression remains real.

Check for yourself how the expression simplifies for pure real or pure imaginary roots.