I have a 2nd order differential equation here. The main thing I need help for is working out what the general solution will take the form of. The equation I have is: $$ \ddot{x}+2\xi\omega\dot{x}+\omega^2x=0$$ I can solve for the constants but I just don't really know what general solution to look for and more importantly how to go about looking for it.

  • $\begingroup$ You "can solve for the constants". What constants? If you show the work you've done and where you are stuck we'll be able to help you understand. $\endgroup$ – user88595 May 22 '14 at 9:17
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    $\begingroup$ That said, the general method is to substitute $x(t) = e^{t\lambda}$ and solve the quadratic in $\lambda$. $\endgroup$ – user88595 May 22 '14 at 9:19
  • $\begingroup$ Are $\omega,\xi$ constants? If so, use $x''=x'\frac{dx'}{dx}$ $\endgroup$ – evil999man May 22 '14 at 9:19

I suppose that $x$ stands for $x(t)$. So, for $$\ddot{x}+2\xi\omega\dot{x}+\omega^2x=0$$ the characteristic equation is, as usual, $$r^2+2\xi\omega r+\omega^2=0$$ for which you get the roots $$r_{1,2}=-\xi \omega\pm \omega\sqrt{\left(\xi ^2-1\right) }$$ and the general solution is $$x(t)=c_1 e^{r_1t}+c_2 e^{r_2t}$$

  • $\begingroup$ Unless, of course, $\xi=\pm1$. $\endgroup$ – Gerry Myerson May 22 '14 at 10:01
  • $\begingroup$ Aren't you tired to be always right ? Thanks for pointing. Cheers. $\endgroup$ – Claude Leibovici May 22 '14 at 10:04

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