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

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



  1. With input y'[x] == 1, the desired output is C[1]->y[x] - x.

  2. 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.






differential-equations







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  • 1




    Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
    – xzczd
    2 hours ago










  • 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
    2 hours ago











  • Yes, and that's the reason I didn't vote for close as duplicate.
    – xzczd
    2 hours ago














up vote
1
down vote

favorite












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



  1. With input y'[x] == 1, the desired output is C[1]->y[x] - x.

  2. 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.






differential-equations







share|improve this question









New contributor




Ilya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 1




    Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
    – xzczd
    2 hours ago










  • 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
    2 hours ago











  • Yes, and that's the reason I didn't vote for close as duplicate.
    – xzczd
    2 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











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



  1. With input y'[x] == 1, the desired output is C[1]->y[x] - x.

  2. 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.






differential-equations







share|improve this question









New contributor




Ilya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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



  1. With input y'[x] == 1, the desired output is C[1]->y[x] - x.

  2. 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.






differential-equations




differential-equations






share|improve this question









New contributor




Ilya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









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edited 24 mins ago









m_goldberg

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asked 2 hours ago









Ilya

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Ilya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.







  • 1




    Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
    – xzczd
    2 hours ago










  • 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
    2 hours ago











  • Yes, and that's the reason I didn't vote for close as duplicate.
    – xzczd
    2 hours ago












  • 1




    Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
    – xzczd
    2 hours ago










  • 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
    2 hours ago











  • Yes, and that's the reason I didn't vote for close as duplicate.
    – xzczd
    2 hours ago







1




1




Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
– xzczd
2 hours ago




Strongly related, if not duplicate: mathematica.stackexchange.com/q/137598/1871
– xzczd
2 hours ago












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
2 hours ago





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
2 hours ago













Yes, and that's the reason I didn't vote for close as duplicate.
– xzczd
2 hours ago




Yes, and that's the reason I didn't vote for close as duplicate.
– xzczd
2 hours ago










1 Answer
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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.






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    up vote
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    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.






    share|improve this answer
























      up vote
      2
      down vote













      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.






      share|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        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.






        share|improve this answer












        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        AccidentalFourierTransform

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