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Say I have the 2-D system of ODEs as such:

$x' = y$

$y' = -2\left[Z(x) + y^2\right]y - Z'(x)$

and I wanted to write this as a single ODE, even though it won't help me at all in solving this system, I know! Is it valid to write:

$x = \int y dt$,

and then sub this into the $y'$ ODE and write:

$y' = -2\left[Z(\int y dt) + y^2\right]y - Z'(\int y dt)$

, so that you have one ODE in terms of one variable, $y$ in this case, where $Z$ is some function.

Thank you.

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What about replacing $y$ by $x'$ in the second equation? We thus have $$ x'' = -2 \, (Z(x) + {x'}^2)\, x' - Z'(x) \, , $$ which is a nonlinear second-order ODE. Otherwise, consider the vector $X = (x,y)^\top$. Then, $X$ satisfies the ODE system $$ X' = f(X) \, , $$ where $f(X) = \left(y, -2 \, (Z(x) + y^2)\, y - Z'(x)\right)^\top $.

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  • $\begingroup$ Yes, but for "personal" reasons, I would like to leave everything as a first-order equation, and thus, would like to know if my $y'$ above is valid. $\endgroup$ – Thomas Moore Sep 9 '17 at 17:16
  • $\begingroup$ But the original system is already a system. $\endgroup$ – Thomas Moore Sep 9 '17 at 17:21
  • $\begingroup$ Okay, that's really innovative, thanks. But, is my $y'$ equation in the last line, correct as is? $\endgroup$ – Thomas Moore Sep 9 '17 at 17:24
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    $\begingroup$ @ThomasMoore Yes, but in this form, it is an integro-differential equation, not an ODE. $\endgroup$ – Harry49 Sep 9 '17 at 17:26
  • $\begingroup$ Okay, that rings a bell! Thank you. $\endgroup$ – Thomas Moore Sep 9 '17 at 17:27
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It is almost valid. You just need to add the integration constant and make sure the integral has bounds from some fixed number up to t, where t is the independent variable. A small amount of care should be taken to make sure that the integral exists.

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