From the following equation, assume that the cross-sectional area A is a constant. = (1/m) hc(T-Tfluid) Also assume that all other physical parameters k, h, A, C, L, Twall and T fluid are well known and assumed to be positive constants. (a) Based on these assumptions, the above ODE can be simplified to which of the follow? hc(T-Tfluid) Ока d²T dx2 (k.. dA dk dT + A. = dx dx dx hc(T-T fluid) dT ка = (hC(T-T fluid)) dx dx (k.. dA dk d²T + A. dx dx dx² hc(T-T fluid) d dx (kA) = (hC(TT fluid)) dx ат (b) Solve the ODE based on the following two boundary conditions. For simplicity, let a² T(x) = At x = 0, T = T wall At x = L, dT = 0. dx = Ch ка for when formulating your solution. (Use T, to represent T fluid and Tw to represent T wall in your response.) (c) Define a "dimensionless" variable @(x) = T(X) - Tfluid Twall - Tfluid ee Based on the T(x) solution, rewrite the solution in terms of (x) and the hyperbolic cosine function, where cosh(y) = Again, let a² = 2 8(x) = Ch for when formulating your solution. ка