# Finding the general solution of this ODE

1. Sep 1, 2005

### Benny

Hi can someone help me out with the following question?

Q. The differential equation for unforced under-damped oscillatory motion can be written $$\mathop x\limits^{..} + 2p\mathop x\limits^. + \omega ^2 x = 0$$ where the constants p and omega satisfy $$0 < p < \omega$$.

Find the general solution of this differential equation, and show that this solution can be expressed in the form $$x = {\mathop{\rm Re}\nolimits} ^{ - pt} \cos \left( {\sqrt {\omega ^2 - p^2 } t - \alpha } \right)$$ where R and alpha are constants.

$$x\left( t \right) = e^{ - pt} \left( {c_1 \cos \left( {t\sqrt {p^2 - \omega ^2 } } \right) + c_2 \sin \left( {t\sqrt {p^2 - \omega ^2 } } \right)} \right)$$ by using 0 < p < omega.

As can be seen I haven't been able to get too far. I can't think of a way to get to the answer. Can someone please help me out?

2. Sep 1, 2005

### HallsofIvy

Yes, the characteristic equation is r2+ 2pr+ &omega;= 0 (You have &omega;2 in the differential equation but your solution seems to be for &omega; only) and the solutions to that are $-p\pm\sqrt{p^2-\omega^2}= -p\pm\sqrt{\omega^2-p^}i$.
Yes, the general solution to the differential equation is
$$e^{-pt}\left(C_1 Cos(\sqrt{\omega^2-p^2}t)+C_2 Sin(\sqrt{\omega^2-p^2}t\right)$$.

Now, all you want to do is show that that can be expressed as a single cosine.

Use the trig identity: Cos(a+ b)= Cos(a)Cos(b)- Sin(a)Sin(b).

Obviously, you want $b= \sqrt{\omega^2-p^2}$. You will need to find a such that Cos(a)= C1, Sin(a)= C2. Obviously that can't happen unless the sum of their squares is equal to 1: that's where R comes in: What is the sum of the squares now?