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$ (M+2m)\ddot{x} + m(l_1 \ddot{\theta}_1 \cos\theta_1 - l_1\dot{\theta}_1^2 \sin\theta_1) + m(l_2\ddot{\theta}_2 \cos\theta_2-l_2\dot{\theta}_2^2 \sin\theta_2) = F $

$ l_1\ddot{\theta}_1 + \ddot{x} \cos\theta_1 - g \sin\theta_1 = 0 $

$ l_2\ddot{\theta}_2 + \ddot{x} \cos\theta_2 - g \sin\theta_2 = 0 $

Here can I use Cramer's rule to find $\ddot{x}\ or\ \ddot{\theta}_1 \ or \ \ddot{\theta}_2 $ in terms of the $\dot{\theta}_1,\dot{\theta}_2 \ and \ \dot{x}$ ?

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Yes, of course! (Since then you're just solving an ordinary (algebraic) linear system.) – Hans Lundmark Oct 30 '12 at 8:06
Your system is not linear in those variables (you have some $\dot{\theta}^2$ terms), so Cramer's rule is not applicable. – Christopher A. Wong Oct 30 '12 at 8:12
Yes, of course. – André Nicolas Oct 31 '12 at 2:55
It seems there is a contradiction here... :-) I figured out that, after I solved all these equations I found that Cramer's rule is possible, because my solution in fact was correct... – kemal acikgoz Oct 31 '12 at 13:47

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