How to obtain the solution of an ODE in implicit form?

Question

I want to get the general solution of a first-order ODE in implicit form. It should be something like this:

With input y'[x] == 1, the desired output is C[1]->y[x] - x.
With input y'[x] == 1/y[x]^2 (nonlinear ODE), the desired output is C[1]->y[x]^3/3 - x

DSolve tries to evaluate the explicit form of y[x] by default. Is it possible to keep the implicit solution?

I tried explicit equation integration using Integrate and tracing (Trace with TraceInternal -> True). Neither helped me with this problem.

Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871 — xzczd, Commented Oct 14, 2018 at 12:48
But this seems not to work correctly. Quiet@Trace[DSolve[y'[x] == 1, y[x], x], Solve[e_, y[x]] -> (eqn = e), TraceInternal -> True]; eqn returns -1 + y[x] == 0 with no integration constant — Ilya, Commented Oct 14, 2018 at 12:54
Yes, and that's the reason I didn't vote for close as duplicate. — xzczd, Commented Oct 14, 2018 at 13:04

AccidentalFourierTransform · Accepted Answer · 2018-10-14 14:05:53Z

4

The following works for the two examples in the OP:

eq = y'[x] == 1; (* try also eq = y'[x] == 1/y[x]^2 *)
Solve[Equal @@ DSolve[eq, y[x], x][[1, 1]], C[1]]
(* C[1] -> -x + y[x] *)

Higher order ODEs contain more constants of integration, so OP shall modify the code accordingly.

answered Oct 14, 2018 at 14:05

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$\begingroup$ Thanks a lot! This works perfectly for both cases $\endgroup$
– Ilya
Commented Oct 14, 2018 at 15:18
$\begingroup$ @Ilya I'm glad I could help. If you try this with more examples, and one doesn't work, let me know and I'll see if I can do something. Cheers! $\endgroup$
– AccidentalFourierTransform
Commented Oct 14, 2018 at 15:51

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