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  • $\begingroup$ I wonder what form has that f in the "general" solution ... Can you share some info about it? $\endgroup$ Commented Mar 13, 2015 at 12:40
  • $\begingroup$ @belisarius f is an arbitrary function. In other words, no matter what f you choose, $u \left( x,y\right)$ is a solution to the PDE. To simplify the problem, you can ignore the $f$ function and just use what is inside. $\endgroup$ Commented Mar 13, 2015 at 12:43
  • $\begingroup$ @belisarius Linear and quasilinear first-order PDEs have such a solution. In DSolve[D[u[x, y], x] + D[u[x, y], y] == 1, u, {x, y}] and DSolve[D[u[x, y], x] + u[x, y] D[u[x, y], y] == 1, u, {x, y}] the f shows up as C[1]. $\endgroup$ Commented Mar 13, 2015 at 13:52
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    $\begingroup$ @belisarius I just learned that a few months ago myself :) -- never had a course in PDEs. I think of it like this: An ordinary antiderivative is defined up to a function of no variables (i.e. a constant); an anti-partial-derivative is defined up to a function n-1 variables, where n is the number of variables in the fn./PDE being integrated. $\endgroup$ Commented Mar 13, 2015 at 14:07
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    $\begingroup$ OK, it is confirmed as a bug by Wolfram technical support. $\endgroup$ Commented Mar 17, 2015 at 16:11