I am looking at the following variation of a integro-differential, with y[0]=1. The output is not great, any solutions to this?

Clear[s, t, J, w];
J[w_] := 1/(2 + w^2)
eqn = y'[t] == -Integrate[y[t1] J[w] Exp[I (0.1 - w) (t1 - t)], {w, 0, ∞}, {t1, 0,t}] 

LaplaceTransform[eqn, t, s] 

InverseLaplaceTransform[%, s, t] 

ySol[t_] = y[t] /. First[%] /. y[0] -> 1

Plot[ySol[t], {t, 0, 10}, PlotRange -> {-1, 1}]


I received some feedback from a Wolfram Support Engineer regarding this question — it appears that current versions of Mathematica are unable to solve this sort of integro-differential equations. Hopefully we will be able to see better implementations and future versions can crack this sort of equations.

  • 1
    $\begingroup$ I guess you're trying to use the method I posted here, but the first thing one needs to correct then is that you have to say Solve[%, LaplaceTransform[y[t], t, s]] before doing InverseLaplaceTransform[%, s, t]. That's not the real problem, though. Since you have two integrals, the solution would involve two Laplace transforms. Note that the w integral is itself a Laplace transform integral. $\endgroup$ – Jens May 31 '13 at 5:35
  • $\begingroup$ That's exactly right, this prob. is an extension of the earlier one, however you may note that the integrand involving J(w) can be sorted out at the earlier stage, and then you ony have to deal with one laplace transform? $\endgroup$ – thils May 31 '13 at 5:47
  • $\begingroup$ Come to think of it, I may not be right as the inner inregrant is dependent on y.... $\endgroup$ – thils May 31 '13 at 5:56
  • $\begingroup$ Yes, it's not so simple. One can simplify things with a translation of the t variables, but I'm out of time for today. $\endgroup$ – Jens May 31 '13 at 6:05
  • $\begingroup$ I get this output: "Unable to prove that integration limits {0,t} are real. Adding \ assumptions may help" $\endgroup$ – thils May 31 '13 at 6:54

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