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- Thread starter dwsmith
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- #2

- Jan 26, 2012

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\begin{align*}

x_{1}&=x \\

x_{2}&= \dot{x} \\

y_{1}&= \theta \\

y_{2}&= \dot{ \theta} \\

0&=L \dot{y}_{2}+x_{2} y_{2}+ \dot{x}_{2} y_{1}.

\end{align*}

There is no $x$ in the original ODE, so you could theoretically integrate that once immediately by essentially leaving out the first equation.

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- #3

- Feb 7, 2012

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Notice that $\dot{x}\dot{\theta} + \ddot{x}\theta = \frac d{dt}(\dot{x}\theta)$, so (assuming that $L$ is a constant) the equation can be written $\frac d{dt}(L\dot{\theta} +\dot{x}\theta) = 0$. You can integrate this once, to get $L\dot{\theta} +\dot{x}\theta = $ const. But you still have the situation of two dependent variables and only one equation, so I don't see how you can go beyond there without further information.Is there a way to break this up into a system of ODEs?

$$

L\ddot{\theta} + \dot{x}\dot{\theta} + \ddot{x}\theta = 0

$$

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I know that acceleration is positive and constant and velocity is positive. Does that offer enough information?Notice that $\dot{x}\dot{\theta} + \ddot{x}\theta = \frac d{dt}(\dot{x}\theta)$, so (assuming that $L$ is a constant) the equation can be written $\frac d{dt}(L\dot{\theta} +\dot{x}\theta) = 0$. You can integrate this once, to get $L\dot{\theta} +\dot{x}\theta = $ const. But you still have the situation of two dependent variables and only one equation, so I don't see how you can go beyond there without further information.