I am tasked with solving this PDE using separation of variables: ($ \alpha , \beta , \gamma $ are constants)

$$ z_{xy} + \alpha z_x + \beta z_y - \gamma z = 0 $$

By assuming $ z = X(x)Y(y) $ with separation constant $ \lambda $, I acquired ODEs:

$$ X' - \lambda X = 0 $$ $$ Y' + \frac{\alpha - \lambda \gamma}{1 + \lambda \beta}Y = 0 $$

These are first order ODEs with solutions:

$$ X = C_{1}\exp(\lambda x) $$ $$ Y = C_{2}\exp\left(\frac{\lambda \gamma - \alpha}{1 + \lambda \beta} y\right) $$

This implies that:

$$ Z = XY = C\exp\left(\lambda x + \frac{\lambda \gamma - \alpha}{1 + \lambda \beta} y\right) $$

However, a second order PDF should have two unknowns and not one. What am I doing wrong?


I think the problem is simply that it is not true, in general, that a solution to a second order PDE should have two arbitrary constants.

For an $n$-th order ODE, it is true that a solution should have $n$ arbitrary constants, but the situation is not the same for an $n$-th order PDE. See for more details the answer here.

  • $\begingroup$ Aren't both C and $\lambda$ undermined constants? $\endgroup$ – user247327 Jan 14 at 19:42
  • $\begingroup$ Yes, but in my experience, when solving a PDE with that method, the separation constant is generally not seen the same way as an integration constant. Since the OP saw only one unknown constant, I assumed that the separation constant was not to be seen as undetermined. In any case, it remains true that one should not seek two undetermined constant when solving a second order PDE. $\endgroup$ – Vincent Jan 14 at 19:51
  • $\begingroup$ Generally, the general solution to a second order PDE will involve two undetermined functions. $\endgroup$ – user247327 Jan 15 at 1:42

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