I am trying to gain some information on a non-linear partial differential equation by issuing the following Mathematica command:

DSolve[ f[x, y] - x D[f[x, y], x, y] == 0 , f[x, y] , {x, y} ]

As an answer I just get the same expression. Does that mean that Mathematica cannot come up with any useful information regarding the solution of such equations?

Are there other things I can try? Is anything known about such differential equations?

Thanks in advance,


P.S. The function Exp[ 2 Sqrt[ y Log[x] ] ] is "almost" a solution of this differential equation.

  • $\begingroup$ this pde is linear. Why do you call it non-linear? $\endgroup$
    – Nasser
    Aug 10, 2023 at 6:49
  • $\begingroup$ The second derivative is multiplied by x, which makes it non-linear. $\endgroup$
    – Uri Z
    Aug 10, 2023 at 6:51
  • $\begingroup$ No. that is wrong. Nonlinearity is on the dependent variable, not the independent variables. The dependent variable enters the pde as linear. $\endgroup$
    – Nasser
    Aug 10, 2023 at 6:53
  • $\begingroup$ Thanks for correcting me. Is there anything you can say regarding the queation itself? $\endgroup$
    – Uri Z
    Aug 10, 2023 at 7:40
  • $\begingroup$ By $ x \partial_x \to \partial_z $ it is the wave equation $\partial_{z,y}f =f$. $\endgroup$
    – Roland F
    Aug 10, 2023 at 11:37

1 Answer 1


Mathematica can't solve this directly. One way around is to help it a little by using separation of variables. By assuming $f(x,y)= X(x) Y(y)$

pde=f[x,y]-x D[f[x,y],x,y]==0

Mathematica graphics

The above shows that we have

$$ x \frac{X'}{X} \frac{ Y'}{Y} = 1 $$ Which is the same as $$ x \frac{X'}{X} = \frac{ Y}{Y'} $$ Hence these both are same and equal to some constant, say $\lambda$ which is the eigenvalue. $$ x \frac{X'}{X} = \frac{ Y}{Y'} = \lambda $$

So we have two ode's to solve

solX  =  DSolveValue[ode1,X[x],x]
solY  =  DSolveValue[ode2,Y[y],y]/.C[1]->C[2]

Hence the final solution is


Mathematica graphics

The eigenvalue $\lambda$ can be determined from boundary conditions.


To verify that the above solution is correct

sol=f->Function[{x,y},E^(y/λ) x^λ C[1] C[2]]
pde=f[x,y]-x D[f[x,y],x,y]==0

Mathematica graphics

  • $\begingroup$ Thanks. That's interesting! As I said in the post, the function $e^{\sqrt{y \ln x}}$ is "close" to being a solution of the equation. I wonder whether there is an exact solution of a similar form. Such a solution, if it exists, is not separable. Is anything known about non-separable solutions? $\endgroup$
    – Uri Z
    Aug 10, 2023 at 7:51
  • $\begingroup$ The more general solution is Integrate[c[λ] E^(y/λ) x^λ, λ], where c is an arbitrary function of λ. $\endgroup$
    – bbgodfrey
    Aug 11, 2023 at 1:36

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