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Consider $a\ddot{x}+ib\dot{x}+cx=0$. Let's find solutions of the form $x(t)=Ae^{rt}$.

One writes down the characteristic equation: $ar^2+ibr+c=0$, with solutions

$$r_{1,2}=\frac{-ib \pm \sqrt{b^2-4ac} }{2a}.$$

Further suppose the problem states $4ac>b^2$, such that $\sqrt{b^2-4ac}=i\sqrt{4ac-b^2}$. Then, our roots are fully complex:

$$r_{1,2}=i\ \frac{-b \pm\sqrt{4ac-b^2} }{2a}.$$

So we obtain oscillatory solutions with frequencies $w_{1,2}=-i\ r_{1,2}$.

Now, let's say we knew apriori that the solutions are oscillatory, and instead considered a form $x(t)=Ae^{iwt}$. Plugging into the differential equation, we obtain

$$aw^2+bw-c=0 \ \hspace{1cm}\Rightarrow\hspace{1cm} w_{1,2}=\frac{-b\pm\sqrt{b^2+4ac}}{2a}.$$

I don't understand why we do not recover the same frequencies we found when considering the more general $x(t)=Ae^{rt}$ solution. To me it looks like our first approach is a more general one that should account fully for our second approach.

Do complex coefficients alter the applicability of the characteristic equation? How does one explain the inconsistency here?

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  • $\begingroup$ Your discriminant should be $-b^2-4ac$. And the equation should be $a\omega^2-b\omega-c=0$. $\endgroup$
    – user65203
    Commented Apr 1, 2021 at 20:51
  • $\begingroup$ Yikes, you're so right, the discriminant is $-b^2-4ac$. I must have just taken the magnitude by mistake. Although I don't think your second equation is correct. If I fix my first discriminant, the inconsistency goes away. $\endgroup$
    – Ptheguy
    Commented Apr 1, 2021 at 20:58
  • $\begingroup$ You are right about the equation, my bad. $\endgroup$
    – user65203
    Commented Apr 2, 2021 at 7:14

1 Answer 1

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It's $\Delta= -(b^2+4ac)=i^2(b^2+4ac)$ and you can also use a change of variable: $$a\ddot{x}(t)+ib\dot{x}(t)+cx(t)=0$$ Substitute $y=it$ $$i^2a\dfrac {d^2x}{dy^2}+i^2b\dfrac {dx}{dy}+cx=0$$ $$a\dfrac {d^2x}{dy^2}+b\dfrac {dx}{dy}-cx=0$$ Solve and substitute back $y=it$.

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