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Breaking up an ODE

dwsmith

Well-known member
Feb 1, 2012
1,673
Is there a way to break this up into a system of ODEs?
$$
L\ddot{\theta} + \dot{x}\dot{\theta} + \ddot{x}\theta = 0
$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
It's a little unusual that you have a single second-order ODE in two independent variables, but you could just do this:
\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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,712
Is there a way to break this up into a system of ODEs?
$$
L\ddot{\theta} + \dot{x}\dot{\theta} + \ddot{x}\theta = 0
$$
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
I know that acceleration is positive and constant and velocity is positive. Does that offer enough information?