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I am looking to solve the following equations numerically:

$a x=\frac{d}{dt}\left(f(x,y,t)\frac{dy}{dt}\right),\quad b y=\frac{d}{dt}\left(g(x,y,t)\frac{dx}{dt}\right)$

For arbitrary functions $f$ and $g$ and constants $a$ and $b$. I am struggling to find a way to transform this into a system of first order differential equations that I can pass into a solver. It looks like I will need to define these implicitly, but I'm not sure how to do that.

My best attempt so far is the following:

\begin{align}z_1&=f(x,y,t)\frac{dy}{dt}\\ z_2&=g(x,y,t)\frac{dx}{dt}\\ z_3&=ax\\ z_4&=by\\ \end{align}

\begin{equation}\begin{pmatrix}z_1\\z_2\\z_3\\z_4\end{pmatrix}'=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&\frac{a}{g(t,x,y)}&0&0\\\frac{b}{f(t,x,y)}&0&0&0\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\\z_4\end{pmatrix} \end{equation}

However, this seems fairly inelegant and assumes that you are always able to divide by $f$ and $g$. I'm trying to keep this as general as possible, so don't want to make that assumption. Is there a better way to turn this into a system of first order differential equations implicitly? Thanks!

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  • $\begingroup$ You always have to exclude $f=0$ and $g=0$ from the domain of the ODE system, as on those surfaces the order of the system collapses, the system becomes singular there. Now you could ask under what circumstances a solution can be prolonged into these singular sets and continued (uniquely?) on the other side. $\endgroup$ – LutzL Feb 21 at 9:57
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I would rather define $u = x'$ and $v = y'$. Then, your system becomes \begin{equation*} \begin{pmatrix} x'\\y'\\fv'\\gu' \end{pmatrix} = \begin{pmatrix} u\\v\\ ax -f_xuv -f_yv^2\\ by -g_xu^2 - g_yuv \end{pmatrix} \end{equation*}

As long as $f\neq 0\neq g$, this is a nonlinear system of first order ODEs. If, at a certain point, $f=0$, or $g=0$, or both, then this system is a differential-algebraic equation (DAE, see here and here). Numerical methods for DAEs are well studied and available. For instance, see Matlab built-in function.

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