# 2nd Order Differential Equation - general solution

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.

• 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. – user88595 May 22 '14 at 9:17
• That said, the general method is to substitute $x(t) = e^{t\lambda}$ and solve the quadratic in $\lambda$. – user88595 May 22 '14 at 9:19
• Are $\omega,\xi$ constants? If so, use $x''=x'\frac{dx'}{dx}$ – 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 $$x(t)=c_1 e^{r_1t}+c_2 e^{r_2t}$$
• Unless, of course, $\xi=\pm1$. – Gerry Myerson May 22 '14 at 10:01