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I am trying to find non-trivial solutions of the ODE $\lambda u'(x) = u(x+R) - u(x)$ using Mathematica.

In particular, I am using the command 

 DSolve[a*y'[x] == y[x + r] - y[x], y[x], x]

which for a standard (i.e., non-delayed, non-anticipated ODE) works fine.

In this case I got the error

DSolve::litarg: "To avoid possible ambiguity, the arguments of the dependent variable in !({a\\ *SuperscriptBox[\"y\", \"[Prime]\", MultilineFunction->None][x] == (-y[x]) + y[r + x]}) should literally match the independent variables."

Any advice?

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  • $\begingroup$ Have you read the documentation yet? $\endgroup$
    – Mr.Wizard
    Commented Dec 23, 2012 at 16:55
  • $\begingroup$ @Mr.Wizard, sorry, thanks for the doc link. Indeed I googled the error; although I found general answers which I am too newbie o be able to uderstand. Be patient. $\endgroup$
    – Acorbe
    Commented Dec 23, 2012 at 17:00
  • $\begingroup$ @Mr.Wizard that doc page doesn't really help solve the DDE $\endgroup$
    – acl
    Commented Dec 23, 2012 at 17:04
  • $\begingroup$ I suggest you look at this doc page $\endgroup$
    – m_goldberg
    Commented Dec 24, 2012 at 0:56

1 Answer 1

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This is a delay differential equation, not an ODE. Mathematica can numerically solve DDEs with constant delays, eg, scaling u and x to reduce your DDE to $u'(x)=u(x+1)-u(x)$, we can do

sln = NDSolve[{u'[x] == u[x - 1] - u[x], 
       u[x /; x <= 0] == x^2}, u, {x, -1, 5}];
Plot[u[x] /. sln, {x, -1, 1}]

Mathematica graphics

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  • $\begingroup$ No way to have some sort of general integral, I guess, right? $\endgroup$
    – Acorbe
    Commented Dec 23, 2012 at 16:58
  • $\begingroup$ not that I know of $\endgroup$
    – acl
    Commented Dec 23, 2012 at 16:58
  • $\begingroup$ Thanks for the precious help, btw. $\endgroup$
    – Acorbe
    Commented Dec 23, 2012 at 20:01

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