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Finding the general solution of this ODE

  • Thread starter Benny
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Hi can someone help me out with the following question?

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

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

[tex]
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)
[/tex] 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?
 

HallsofIvy

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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 [itex]-p\pm\sqrt{p^2-\omega^2}= -p\pm\sqrt{\omega^2-p^}i[/itex].
Yes, the general solution to the differential equation is
[tex]e^{-pt}\left(C_1 Cos(\sqrt{\omega^2-p^2}t)+C_2 Sin(\sqrt{\omega^2-p^2}t\right)[/tex].

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 [itex]b= \sqrt{\omega^2-p^2}[/itex]. 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?
 

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