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I am new to Mathematica in general. Just trying to solve a couple of coupled ODES. Dont really understand the error it throws.

DSolve[{y'[t] == A*x[t]/((x[t]^2 + y[t]^2)*(1 - R/Sqrt[(x[t]^2 + y[t]^2)])), 
  x'[t] == A*y[t]/((x[t]^2 + y[t]^2)*(1 - R/Sqrt[(x[t]^2 + y[t]^2)])),
   x[0] == 1, y[0] == 1}, {x, y}, t, 
 Assumptions -> x \[Element] Reals, y \[Element] Reals].

This gives the error "DSolve::derlen: The length of the derivative operator Derivative[1] in y'[t] is not the same as number of arguements." Now this system actually has a singularity if x^2+y^2=R. I just don't understand how to put in that as an assumption. I have tried "Assumptions-> x[t]^2+y[t]^2<R", it doesnt work. Moreover I have a physical understanding that if the system starts at x[0]=1 and y[0]=1 it should not reach this singularity since the solution would go in a circle around the centre. I might have made some superflous errors as well, so I am grateful for any help

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    $\begingroup$ Put braces {} around assumptions: Assumptions -> {…..} $\endgroup$
    – Michael E2
    Commented Jun 24, 2021 at 1:28
  • $\begingroup$ You might want to report this to WRI as a bug. It probably ought to do a better job of parsing here, because an option precedes the Element[y, Reals] argument that looks like a domain spec. It’s an unhelpful error message in any case. $\endgroup$
    – Michael E2
    Commented Jun 24, 2021 at 1:32
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    $\begingroup$ Miraculously, DSolve can find the general solution (remove the initial conditions). You’ll have to solve for the integration parameters yourself. The solutions is rather complicated and it might be difficult. $\endgroup$
    – Michael E2
    Commented Jun 24, 2021 at 1:42

1 Answer 1

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Try “homogeneous” initial conditions:

DSolve[{y'[t] == A*x[t]/((x[t]^2 + y[t]^2)*(1 - R/Sqrt[(x[t]^2 + y[t]^2)])), 
  x'[t] == A*y[t]/((x[t]^2 + y[t]^2)*(1 - R/Sqrt[(x[t]^2 + y[t]^2)])),
   x[0] == a*R, y[0] == b*R}, 
 {x, y}, t]
(* long output *)

It gives two solutions, perhaps from rationalizing the square roots. You should check which satisfy a particular initial conditions.

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  • $\begingroup$ I tried it with some numerical value of R and A,with a,b<1 and x,y=Real, then it gives errors which state "For some branches of the general solution, the given boundary \ conditions lead to an empty solution" and "Unable to resolve some of the arbitrary constants in the general \ solution using the given boundary conditions. It is possible that \ some of the conditions have been specified at a singular point for \ the equation.". Which is interesting because if it starts from some radius lower than R and x and y are real then it should not hit singularity. $\endgroup$
    – Sarthak Choudhury
    Commented Jun 24, 2021 at 23:03
  • $\begingroup$ @SarthakChoudhury The boundary conditions are not easy equations to solve symbolically. I would check the DSolve solution from my code against NDSolve for numerical values of R etc. You can plug values in with mysol /. {R -> 2} and so forth. I wouldn't plug the numerical values into the ODEs and call DSolve again. (I guess I might try it, but I'm not surprised DSolve had trouble.) $\endgroup$
    – Michael E2
    Commented Jun 24, 2021 at 23:27
  • $\begingroup$ Thanks a lot @Michael E2. I really appreciate your help $\endgroup$
    – Sarthak Choudhury
    Commented Jun 25, 2021 at 1:52

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