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I have two supposedly homogeneous equations. Solve gives a trivial solution for them. Is there any way in Mathematica to extract non-trivial solutions for this system. The equations are:

eq1 = (1/(NTUC^2 + 4 m^2 π^2)) l L (C1nm (2 (-1 + E^NTUC) NTUC^3 pc + 4 E^NTUC m^2 NTUC^2 π^2 (pc - 2 γ) + 16 E^NTUC m^4 π^4 (pc - γ) - E^NTUC NTUC^4 γ) + C2nm (2 (-1 + E^NTUC) NTUC^3 pc + E^NTUC NTUC^4 γ + 16 E^NTUC m^4 π^4 (pc + γ) + 4 E^NTUC m^2 NTUC^2 π^2 (pc + 2 γ))) == 0


eq2 = (1/(NTUH^2 + 4 n^2 π^2)) l L (C2nm (2 (-1 + E^NTUH) NTUH^3 ph + 4 E^NTUH n^2 NTUH^2 π^2 (ph - 2 γ) + 16 E^NTUH n^4 π^4 (ph - γ) - E^NTUH NTUH^4 γ) + C1nm E^(2 w γ) (2 (-1 + E^NTUH) NTUH^3 ph + E^NTUH NTUH^4 γ + 16 E^NTUH n^4 π^4 (ph + γ) + 4 E^NTUH n^2 NTUH^2 π^2 (ph + 2 γ))) == 0

solnm = Solve[{eq1,eq2}, {C1nm, C2nm}]

The variables to be solved for are C1nm, C2nm. The last command returns C1nm->0,C2nm->0. What are my alternatives to get non-trivial solutions for this system ? Can numerical methods work ?

Some constant values for the interested user:

n,m can take all positive integer values not equal to zero.

Also , γ = Sqrt[(2 n π/L)^2 + (2 m π/l)^2]

L = 0.025; l = 0.025; w = 0.003; NTUH = 0.433; NTUC = 0.433; ph = 65.24; pc = 65.24;
thi = 120; tci = 30;
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    $\begingroup$ If I make the numerical replacements from above and choose m=1 and n=2 I get a system of equations that has the form: m.{C1nm, C2nm} =={0,0}. As the determinant of m is not equal to zero, this system has only the trivial solution as MMA indicates. $\endgroup$ – Daniel Huber Sep 23 at 9:17

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