Finding unkown function

Question

I want to find a function t3

t3[x1, x2, x3] 

that fulfills these equations:

2 D[t3[x1, x2, x3], x3] + 6 D[t3[x1, x2, x3], x2] - 9 x2^2 D[t3[x1, x2, x3], x1] !=0

for any real x

D[t3[x1, x2, x3], x3] + 3 D[t3[x1, x2, x3], x2] + 
    (-1 - 9/2 x1^2) D[t3[x1, x2, x3], x1] == 0

How can I do this (with Mathematica)?

A simpler example

Answer 1

Let's first solve the equation:

sol[x1_, x2_, x3_] = 
 t3[x1, x2, x3] /. 
  DSolve[D[t3[x1, x2, x3], x3] + 
       3 D[t3[x1, x2, x3], x2] + (-1 - 9/2 x1^2) D[t3[x1, x2, x3], x1] == 0, 
        t3[x1, x2, x3], {x1, x2, x3}][[1]]

C[1] here is an arbitrary function of two variables.

Let's take the second condition:

sol2[x1_, x2_, x3_] = 
 FullSimplify[
  D[sol[x1, x2, x3], x3] + 6 D[sol[x1, x2, x3], x2] - 9 x2^2 D[sol[x1, x2, x3], x1]
 ]

Now, I'm not sure what is != supposed to mean: for the expression to not be identically equal to zero, or to never attain the value zero? If the former, then with

f[a_, b_] := a^2 + b^2

for example

sol2[1, 1, 1] /. C[1] -> f // N

3.71851

so the expression is not identically zero. Thence, the solution is

sol[x1, x2, x3] /. C[1] -> f

Here

FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> f, {x1, x2, x3}, Reals]

{{x1 -> 0, x2 -> 0, x3 -> 0}}

---

On the other hand, if C[1] == ArcTan, then

sol2[1, 1, 1] /. C[1] -> ArcTan // N

-0.365403

so the solution is not identically equal to zero. Moreover,

FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> ArcTan, {x1, x2, x3}, Reals]

didn't give any instance for several minutes, so it's likely there is none and the function is non-zero everywhere.