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I'm interested in finding the general solution $f(x)$ of the equation: $(1+x)f'(x)+(f(x/2)-f(x))=-1$, $x\ge0$. Can Mathematica help here? For instance, it is easy to verify that the (slightly different) equation $(1+x)f'(x)+2(f(x/2)-f(x))=-1$ is solved by $f(x)=C-x$ where $C$ is an arbitrary constant. Can Mathematica help reach this conclusion?

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  • $\begingroup$ Did you check RSolve? $\endgroup$ Apr 10, 2017 at 1:29
  • $\begingroup$ Yes, to no avail. It doesn't seem to understand derivatives. $\endgroup$
    – Alex
    Apr 10, 2017 at 1:34
  • $\begingroup$ I had no luck with either DSolve or NDSolve. DSolve[(1 + x) f'[x] + f[x/2] - f[x] == -1, f, x] gave DSolve::litarg error. NDSolve[{(1 + x) f'[x] + f[x/2] - f[x] == -1, f[0] == 1}, f, {x, 0, 1}] gave NDSolve::cdelay error, saying "the method currently implemented for delay differential equations does not support delays that depend directly on the time variable or dependent variables." $\endgroup$
    – Chris K
    Apr 10, 2017 at 2:27
  • $\begingroup$ @ChrisK: I had exactly the same issues. $\endgroup$
    – Alex
    Apr 10, 2017 at 15:06

1 Answer 1

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A small number of Taylor coefficients in $x=0$ is enough for FindSequenceFunction to provide the whole Taylor series:

n = 10;
res = First[# /. Solve[Thread[Table[SeriesCoefficient[(1 + x) f'[x] + f[x/2] - f[x],
        {x, 0, i}], {i, 0, n}] == Table[SeriesCoefficient[-1, {x, 0, i}], {i, 0, n}]],
            #] &[Table[Derivative[i + 1][f][0], {i, 0, n}]]]
Sum[x^i/i! FindSequenceFunction[res, i], {i, 0, m}] + C[1]

$C[1] + \sum _{i=0}^m -\frac{x^i \prod _{K[1]=1}^{i-1} -2^{-K[1]} \left(2^{K[1]} K[1]-2^{K[1]}+1\right)}{i!}$

If f is replaced by this series in the equation, it is not hard to see that summing up to some m yields something with the form $c\ x^m$. (First include in each of the sums only the 1 or 2 terms that adds some $x^{m-1}$ and expand to see that $x^{m-1}$ disappears, hence after summing the first $m$ terms, there is only $c\ x^m$ that can have a non-zero coefficient). Now sum to $m$, discard all but the $x^m$ terms and simplify to see that $|c\ x^m|$ equals $$|x|^m\prod _{K[1]=1}^m \frac{2^{-K[1]} \left(2^{K[1]} K[1]-2^{K[1]}+1\right)}{K[1]}$$

The factors in the product are all less than one, so the error of the Taylor series decrease to $0$ as $m\to \infty$ if (and probably only if) $|x|\leq 1$.

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