Maybe there's something I'm missing, but I just realized that if we take a certain general ODE such as y''+k^2 y=0 for example, and assume that in a specific case k=0 the solution is vastly different depending on whether k is set to 0 before or after solving the ODE. The general solution to y''+k^2 y = 0 (with k>0) is y=A cos(kx)+B sin(kx). If we then set k=0 the solution reduces to y=A. Yet, if we set k=0 in the original ODE we get  y''=0, whose solution is  y=Bx+C. Where A, B, and C are constants. I get analogous results for y''-k^2 y = 0. On the other hand, if I start with y''+2b y'+k^2 y = 0, then it seems that setting either b or k to 0, but not both, can be done either in the ODE itself or in the general solution. Hrm, I just had another insight while writing this. The form of solution  is due to the characteristic equation having a repeated real root of r=0. Is this the reason why my initial attempt failed? Because I was setting the parameter to the same value as the eigenvalue of the ODE?