Say I have a DE

$$ -\phi \left(\phi \left(\left(6975 \phi ^2-3704 \phi +160\right) \omega '(\phi )+\phi \left(\left(6975 \phi ^2-4688 \phi +266\right) \omega ''(\phi )+\phi \left(2 \left(675 \phi ^2-518 \phi +32\right) \omega ^{(3)}(\phi )-\phi \omega ^{(4)}(\phi )\right)\right)\right)\right) = 0 $$

and I want a recurrence relation for coefficients of a power series solution, such that the constant term $\omega(0)=1$, how can I do that?

I will change symbols to $t$ and $f$ because they will look nicer here

t[f_] := x D[f,x]
R[x_] := -9 x^3 + 5 x^2 - 4 x
L[f_,x_] := R[x] Nest[t,f,4] + R[x]^2 Nest[t,f,3] + f
diffEqn = Expand[L[h[x],x]]
  • $\begingroup$ Can you at least rewrite your ODE in Mathematica's format? $\endgroup$ – J. M.'s technical difficulties Feb 4 at 16:49
  • $\begingroup$ @J.M.isinlimbo i have edited the post $\endgroup$ – Bernoulli Feb 4 at 16:53
  • 2
    $\begingroup$ Have you seen this? The only change needed in your case would be to use SeriesCoefficient[] instead of Series[], so that you (hopefully) get a DifferenceRoot[] object. $\endgroup$ – J. M.'s technical difficulties Feb 4 at 16:57
  • $\begingroup$ @J.M.isinlimbo it doesn't give me a recurrence relation. If I only have one initial condition f[0]==1, it would say "not a well-formed linear differential equation with initial conditions." $\endgroup$ – Bernoulli Feb 4 at 17:18

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