Converting an equation to the form $ax+by=c$

Question

I have a linear equation that comes as a result of applying orthogonality on a non-homogeneous boundary condition with two unknowns $C_1$ and $C_2$. Can someone help me to express this equation in the form:

$$a(C_1) + b(C_2)=c\tag A$$

The code to derive the equation is as follows:

T[x_, y_, z_] = (C1* E^(γ z) + C2* E^(-γ z))*Sin[(α x/L) + β]*Sin[(δ y/l) + θ] + Ta;
tc[x_, y_] = E^(-bcy/l)*{tci + (bc/l)*Integrate[E^(bc*s/l)*T[x, s, 0], {s, 0, y}]};
bc1 = (D[T[x, y, z], z] /. z -> 0) == pc (T[x, y, 0] - tc[x, y]);
ortheq1 = Integrate[(bc1[[1]] - bc1[[2]])*Sin[(α x/L) + β]*Sin[(δ y/l) + θ], {x, 0, L}, {y, 0, l}, Assumptions -> {C1 > 0, C2 > 0, L > 0, l > 0, α > 0, β > 0, γ > 0, δ > 0, θ > 0, NTUC > 0, pc > 0, Ta > 0, tci > 0}] == 0;

One can also use the following to derive ortheq1 but it gives a longer result

ortheq1 = Integrate[bc1[[1]]*Sin[(α x/L) + β]*Sin[(δ y/l) + θ], {x, 0, L}, {y, 0, l}] == Integrate[bc1[[2]]*Sin[(α x/L) + β]*Sin[(δ y/l) + θ], {x, 0, L}, {y, 0, l}];

It is the ortheq1 that I would like in the form $(A)$.

Answer 1

Assuming that the appearance of List in the definition of tc is not on purpose, i.e.

T[x_, y_, z_] = (C1* E^(γ z) + C2* E^(-γ z))*Sin[(α x/L) + β]*Sin[(δ y/l) + θ] + Ta;
tc[x_, y_] = E^(-bcy/l)*(tci + (bc/l)*Integrate[E^(bc*s/l)*T[x, s, 0], {s, 0, y}]);
bc1 = (D[T[x, y, z], z] /. z -> 0) == pc (T[x, y, 0] - tc[x, y]);
ortheq1 = Integrate[(bc1[[1]] - bc1[[2]])*Sin[(α x/L) + β]*Sin[(δ y/l) + θ], {x, 0, L}, {y, 0, l}, Assumptions -> {C1 > 0, C2 > 0, L > 0, l > 0, α > 0, β > 0, γ > 0, δ > 0, θ > 0, NTUC > 0, pc > 0, Ta > 0, tci > 0}] == 0;

Then you can use CoefficientList

Module[{coeffs = CoefficientList[Subtract @@ ortheq1, {C1, C2}]},
  a[C1_] = coeffs[[2, 1]] C1;
  b[C2_] = coeffs[[1, 2]] C2;
  c = -coeffs[[1, 1]];
]
(Subtract @@ ortheq1) - (a[C1] + b[C2] - c) // Simplify
(* 0 *)