I find that DSolve can solve

DSolve[x*D[u[x, y], {x, 2}] - D[u[x, y], x, y] == 0, u, {x, y}]


DSolve[x*D[u[x, y], {x, 2}] - x *D[u[x, y], x, y] == 0, u, {x, y}]


DSolve[x*D[u[x, y], {x, 2}] - 3*D[u[x, y], x, y] == 0, u, {x, y}]}

but cannot solve

DSolve[x*D[u[x, y], {x, 2}] - a *D[u[x, y], x, y] == 0, u, {x, y}]


DSolve[x*D[u[x, y], {x, 2}] - y *D[u[x, y], x, y] == 0, u, {x, y}]

Can I re-pose the equation to make it work? thanks, GB

MMa 10.3, 11.1

  • 2
    $\begingroup$ DSolve[]'s support for PDE equations is still somewhat limited, so don't be surprised if some things don't work yet.See Results:12000.org/my_notes/pde_in_CAS/maple_2019_and_mma_12/index.htm $\endgroup$ – Mariusz Iwaniuk Jul 10 at 21:14
  • 1
    $\begingroup$ You can break the last PDE down into two successive quadratures: Fold[ Function[{eq, var}, DSolve[eq /. First[#1], var, {x, y}]] @@ #2 &, {{}}, Transpose@{{x*D[v[x, y], {x}] - y*D[v[x, y], y] == 0, D[u[x, y], x] == v[x, y]}, {v, u}}] $\endgroup$ – Michael E2 Jul 11 at 5:15
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Michael E2 Jul 12 at 0:39
  • $\begingroup$ @MichaelE2 Yes, something comes out. Being thick between the ears, I have not been successful in backtesting. It is possible to do so? $\endgroup$ – Gary Bollenbach Jul 13 at 17:31
  • $\begingroup$ @MichaelE2. Now it seems to check. Thanks for passing on this useful technique. $\endgroup$ – Gary Bollenbach Jul 14 at 22:10

Try some Assumptions on the first one.

$Assumptions = a > 0

DSolve[x*D[u[x, y], {x, 2}] - a*D[u[x, y], x, y] == 0, u, {x, y}]
(*{{u -> Function[{x, y}, C[1][a*Log[x] + y]/E^(y/a) + C[2][y]]}}*)

Version 12. The second one is still no go in automatic mode. It is, however, separable and could be solved that way.

  • 1
    $\begingroup$ Assumptions -> a \[Element] Reals && a != 0 also works. $\endgroup$ – Michael E2 Jul 11 at 0:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.