How to find the differential equation that has the solution $u=f(x+\frac{y}{z},y+\frac{x}{z})$? [closed]

Question

I should use partial derivatives to find the differential equation that has the following solution:

$ u=f(x+\frac{y}{z},y+\frac{x}{z})$.

Answer 1

There are infinitely many possibilities, naturally. The simplest one is a linear ansatz, $$ (\alpha\partial_x+\beta\partial_y+\gamma\partial_z) u=0 $$ i.e.,

α D[f[x + y/z, y + x/z], x] + β D[f[x + y/z, y + x/z], y] + γ D[f[x + y/z, y + x/z], z] // Simplify

which yields the equations

{z α + z^2 β - x γ == 0, z^2 α + z β - y γ == 0}

with solution $$ \beta \to -\frac{\alpha (y-x z)}{y z-x},\quad\gamma \to -\frac{\alpha (z^3-z)}{x-y z} $$ Of course, one parameter is arbitrary since the equation is linear. We can take $\alpha=x-yz$ to make the equation look as simple as possible,

(x - y z) D[f[x + y/z, y + x/z], x] + (y - x z) D[f[x + y/z, y + x/z], y] + (z - z^3) D[f[x + y/z, y + x/z], z] // Simplify
(* 0 *)

Final answer is $$ (x-y z)\partial_xu+ (y-x z)\partial_yu+z(1-z^2)\partial_zu=0 $$