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Does this second order differential equation with the time-varying coefficient fit into any general form?

$${d^2x(t)\over dt^2}+\Big(k_1+k_2\cos(\omega t)\Big){dx(t)\over dt}+\Big({1\over 1+k_\Delta \sin(\omega t)}+k_3 \cos(\omega t)\Big) x(t)=F(t)$$ Some properties of the coefficients - $k_1$ and $k_2$ are in similar orders of magnitude and $k_\Delta<<1$.

How can we find the solution of this kind?

Otherwise, how can we analyze the characteristics of a system governed by this? For instance, conditions for damping when we don't have any input excitation i.e., $F(t)=0$, or when we have a harmonic excitation i.e., $F(t)=A \sin(\omega_0 t)$?

Update 1

I understand that if we can get rid of the coefficient in $x'$ term (or remove the time dependency), then the equation can be reduced to a form of the Hill equation. Now, I am trying to do a variable substitution that is similar to this answer.

Update 2

Thanks for the comments and suggestions so far.. From this answer, I understand that having a closed form of a solution would be really challenging (or even impossible). Now, I am trying to do some simplifications and my equation can be reduced to $${d^2x(t)\over dt^2}+\Big(k_1+k_2\cos(\omega t)\Big){dx(t)\over dt}+k_3 x(t)=F(t)$$ It looks much simpler than the initial version, however, still I could not find any general form similar to this. Analytical solvers could not solve it either.

Now, I am only interested in analyzing the characteristics of the system behavior. For example, I can see from numerical simulations that $x(t)$ will have a growing (close to exponential) oscillations when $k_2/k_1>\rm{a~threshold}$, and otherwise stable oscillations. But I could not find such relation analytically.

Any suggestions to move forward from here is highly appreciated.

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    $\begingroup$ Your coefficients are periodic, so Floquet theory applies. Another second order ODE with periodic coefficients is the Hill equation, this might also give qualitative insight. $\endgroup$ – Dr. Lutz Lehmann Feb 11 at 8:12
  • $\begingroup$ @LutzL, Thanks, will look into Floquet theory. I tried to get some qualitative insight from Mathieu function (a variation of Hill equation) but successful so far. Do you have any reference suggestion, which discusses more on the qualitative characteristics of Hill equation? $\endgroup$ – Pojj Feb 11 at 8:24
  • $\begingroup$ @LutzL, from a quick research, I understand that the Floquet theory applies to the differential equations in the form of $x'' + A(t)x=0$ - still in the form of Hill equation. But here, I have non-zero coefficients in $x'(t)$ term. Can I reduce my version into that form? $\endgroup$ – Pojj Feb 11 at 8:31
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    $\begingroup$ In some way, your equation is a perturbation of $x''+k_1x'+x=F$, so that the behavior of that equation should dominate for small $k_2, k_Δ$. I was giving the keywords to be sure you checked these associations, I'm not sure how far they will reach. There is some further theory about averaged behavior, that is, an intelligent way to replace the coefficients with some kind of averages over a period. I'd guess that it leads to the last coefficient being $\frac1{\sqrt{1+k_Δ^2}}$ or similar in the constant-coefficient approximation. $\endgroup$ – Dr. Lutz Lehmann Feb 11 at 8:46
  • $\begingroup$ Thanks, I am trying to get something like this answer to simplify it into that form. $\endgroup$ – Pojj Feb 11 at 11:49
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Hint:

For $\dfrac{d^2x}{dt^2}+(k_1+k_2\cos(\omega t))\dfrac{dx}{dt}+\left(\dfrac{1}{1+k_\Delta\sin(\omega t)}+k_3\cos(\omega t)\right)x=F(t)$ ,

Apply Tangent half-angle substitution:

Let $r=\tan\dfrac{\omega t}{2}$ ,

Then $\dfrac{dx}{dt}=\dfrac{dx}{dr}\dfrac{dr}{dt}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega(r^2+1)}{2}\dfrac{dx}{dr}$

$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}\right)=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d}{dt}\left(\dfrac{dx}{dr}\right)+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d}{dr}\left(\dfrac{dx}{dr}\right)\dfrac{dr}{dt}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d^2x}{dr^2}\dfrac{\omega}{2}\sec^2\dfrac{\omega t}{2}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega^2}{4}\left(\sec^4\dfrac{\omega t}{2}\right)\dfrac{d^2x}{dr^2}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)}{2}\dfrac{dx}{dr}$

$\therefore\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)}{2}\dfrac{dx}{dr}+\left(k_1-\dfrac{k_2(r^2-1)}{r^2+1}\right)\dfrac{\omega(r^2+1)}{2}\dfrac{dx}{dr}+\left(\dfrac{1}{1+\dfrac{k_\Delta r}{r^2+1}}-\dfrac{k_3(r^2-1)}{r^2+1}\right)x=F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

$\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)+(k_1-k_2)\omega r^2+(k_1+k_2)\omega}{2}\dfrac{dx}{dr}+\left(\dfrac{r^2+1}{r^2+k_\Delta r+1}-\dfrac{k_3(r^2-1)}{r^2+1}\right)x=F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

$\dfrac{d^2x}{dr^2}+\left(\dfrac{2r}{r^2+1}+\dfrac{2(k_1-k_2)}{\omega(r^2+1)}+\dfrac{4k_2}{\omega(r^2+1)^2}\right)\dfrac{dx}{dr}+\left(\dfrac{4}{\omega^2(r^2+k_\Delta r+1)(r^2+1)}-\dfrac{4k_3(r^2-1)}{\omega^2(r^2+1)^3}\right)x=\dfrac{4}{\omega^2(r^2+1)^2}F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

For $\dfrac{d^2x}{dt^2}+(k_1+k_2\cos(\omega t))\dfrac{dx}{dt}+k_3x=F(t)$ ,

Apply Tangent half-angle substitution:

Let $r=\tan\dfrac{\omega t}{2}$ ,

Then $\dfrac{dx}{dt}=\dfrac{dx}{dr}\dfrac{dr}{dt}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega(r^2+1)}{2}\dfrac{dx}{dr}$

$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}\right)=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d}{dt}\left(\dfrac{dx}{dr}\right)+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d}{dr}\left(\dfrac{dx}{dr}\right)\dfrac{dr}{dt}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega}{2}\left(\sec^2\dfrac{\omega t}{2}\right)\dfrac{d^2x}{dr^2}\dfrac{\omega}{2}\sec^2\dfrac{\omega t}{2}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega^2}{4}\left(\sec^4\dfrac{\omega t}{2}\right)\dfrac{d^2x}{dr^2}+\dfrac{\omega^2}{2}\left(\sec^2\dfrac{\omega t}{2}\tan\dfrac{\omega t}{2}\right)\dfrac{dx}{dr}=\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)}{2}\dfrac{dx}{dr}$

$\therefore\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)}{2}\dfrac{dx}{dr}+\left(k_1-\dfrac{k_2(r^2-1)}{r^2+1}\right)\dfrac{\omega(r^2+1)}{2}\dfrac{dx}{dr}+k_3x=F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

$\dfrac{\omega^2(r^2+1)^2}{4}\dfrac{d^2x}{dr^2}+\dfrac{\omega^2r(r^2+1)+(k_1-k_2)\omega r^2+(k_1+k_2)\omega}{2}\dfrac{dx}{dr}+k_3x=F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

$\dfrac{d^2x}{dr^2}+\left(\dfrac{2r}{r^2+1}+\dfrac{2(k_1-k_2)}{\omega(r^2+1)}+\dfrac{4k_2}{\omega(r^2+1)^2}\right)\dfrac{dx}{dr}+\dfrac{4k_3x}{\omega^2(r^2+1)^2}=\dfrac{4}{\omega^2(r^2+1)^2}F\left(\dfrac{2\tan^{-1}r}{\omega}\right)$

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  • $\begingroup$ Thanks, but it seems to be even more complicated to move forward from here. Now, we have higher order polynomial fractions in time-varying coefficients. $\endgroup$ – Pojj Feb 18 at 14:31

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