I'm having issues understanding/solving this question I got for an exam:

Consider this nonlinear differential equation (1): $x''(t) + \cos(x(t))\cdot x(t) = \sin(t)$

Suppose $x$ is small and make an expansion of $\epsilon\cdot x$ in (1) of the form $\epsilon\cdot x = x_0\cdot \epsilon + x_1\cdot \epsilon^2 + x_2\cdot \epsilon^3 + ...$

Question: Find the equations for

  • the lowest order of $\epsilon$
  • the second lowest order of $\epsilon$

The equations are NOT supposed to be solved. Any help would be greatly appreciated!

  • $\begingroup$ How have you tried to start? Did you replace $x(t)$ in the problem with $\epsilon x_0(t)+\epsilon^2 x_1(t)$? $\endgroup$ – David Jul 14 '17 at 21:56
  • $\begingroup$ Yes I did. Then I tried to look at the equations I would get for $O(\epsilon^0)$, $O(\epsilon)$, $O(\epsilon^2)$ and so on, but I have no idea if this is the right approach. $\endgroup$ – James Jul 14 '17 at 21:59
  • $\begingroup$ That's the right approach. You can use the Taylor series expansions of $\sin$ and $\cos$ (around $\epsilon=0$, not $t=0$!) to simplify the results as well. $\endgroup$ – David Jul 14 '17 at 22:01
  • $\begingroup$ @David: That $\sin(t)$ is troubling me. Won't the equation at $O(\epsilon^0)$ be $\sin (t) = 0$? I wonder whether there's a typo in the equation. $\endgroup$ – Michael Seifert Jul 14 '17 at 22:03
  • $\begingroup$ Thanks! As far as I can tell, wouldn't the equation with the lowest order of $\epsilon$ be the equation you get for $O(\epsilon)$? $\endgroup$ – James Jul 14 '17 at 22:06

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