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

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

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

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

• And what's the point of answering such irrelevant question, which is not even about Mathematica? Commented Apr 28, 2022 at 11:52
• @yarchik I interpreted the question as: how to solve this using Mathematica. Whether this was actually OP's intention or not does not seem all that relevant to me. Commented Apr 28, 2022 at 11:54
• What is relevant, however, is that the question should show at least a little bit of work. Commented Apr 28, 2022 at 13:02
• @user64494 Indeed, it is not a PDE, it is a trivial identity, just like $x+(-2x)/2=0$. Commented Apr 28, 2022 at 13:53
• You are probably right with (x - y z) D[u[x, y, z], x] + (y - x z) D[u[x, y, z], y] + (z - z^3) D[u[x, y, z], z] == 0. Commented Apr 28, 2022 at 13:56