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I am solving a nonlinear ODE using NDSolve, and one of the terms is an implicit nonlinear function of the dependent variable, for which I am using FindRoot. Here's a trivial example:

y[x_] := y /. FindRoot[Exp[Log[y]] == x, {y, 1}];

I can plot this function, and it gives a straight line y= x, but when I put the function into NDSolve, I get an error message.

s = NDSolve[{h'[t] == -y[h[t]], h[0] == 1}, h, {t, 0, 2}]

and then

Plot[Evaluate[h[t]/.s],{t,0,2}]

If the function y[h[t]] were working, the solution would be h[t]=Exp[-t]. Does every term in NDSolve have to be an explicit function? Is there a way around this?

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    $\begingroup$ Functions that make use of numeric techniques should have their arguments restricted to numeric values: y[x_?NumericQ] := ... $\endgroup$ – Bob Hanlon Jan 21 '18 at 21:44
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  • $\begingroup$ Related: mathematica.stackexchange.com/questions/18393/… $\endgroup$ – Michael E2 Jan 21 '18 at 22:45
  • $\begingroup$ Why not use NDSolve[{Exp[Log[-h'[t]]] == h[t], h[0]==1}, h, {t, 0, 2}] instead? $\endgroup$ – Carl Woll Jan 22 '18 at 16:29
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Simply use ?NumericQ an it works

y[x_?NumericQ] := y /. FindRoot[Exp[Log[y]] == x, {y, 1}];

s = First@NDSolve[{h'[t] == -y[h[t]], h[0] == 1}, h, {t, 0, 2}]

Plot[Evaluate[h[t] /. s], {t, 0, 2}]

Oh, I just see, @Bob Hanlon had the same advice before.

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