Using ./ - having trouble understanding why it isn't more like a 'map' function

*To*: mathgroup at smc.vnet.net*Subject*: [mg124854] Using ./ - having trouble understanding why it isn't more like a 'map' function*From*: trayres <trayres at gmail.com>*Date*: Thu, 9 Feb 2012 05:41:51 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

Hi all- I'm searching through the help for NDSolve, and I run into the following: Plot[Evaluate[y[x] /. s], {x, 0, 30}, PlotRange -> All] Also, in the following tutorial on DSolve Plotting: http://reference.wolfram.com/mathematica/tutorial/DSolvePlottingTheSolution.html the following occurs: Plot[Evaluate[y[x] /. sol /. {C[1] -> 1}], {x, -7, 7}, PlotRange -> All] The trouble is, in the above tutorial I don't understand why you use y[x] /. sol at all. Shouldn't this just be replacing y[x] with sol? If I run 'sol' directly, I get {{y -> Function[{x}, E^(-(x^2/2)) C[1] + 1/2 E^(-(x^2/2)) Sqrt[\[Pi]/ 10] (Sqrt[1 + 2 I] Erfi[Sqrt[1/2 - I] x] + Sqrt[1 - 2 I] Erfi[Sqrt[1/2 + I] x])]}} if I run y[x]/.sol, I get: {E^(-(x^2/2)) C[1] + 1/2 E^(-(x^2/2)) Sqrt[\[Pi]/ 10] (Sqrt[1 + 2 I] Erfi[Sqrt[1/2 - I] x] + Sqrt[1 - 2 I] Erfi[Sqrt[1/2 + I] x])} Then later, if I see what y[x] is now equal to, I get the same thing as y[x]/.sol: {E^(-(x^2/2)) C[1] + 1/2 E^(-(x^2/2)) Sqrt[\[Pi]/ 10] (Sqrt[1 + 2 I] Erfi[Sqrt[1/2 - I] x] + Sqrt[1 - 2 I] Erfi[Sqrt[1/2 + I] x])} This means that /. isn't replacing in an instance, it's actually mapping 'f' (the label for the function) with variable 'x' to the appropriate results of 'sol'. I don't get why Map isn't used, and I don't get what this /. is really doing. If someone could give me a better idea, I'd really appreciate it. Thanks, -Travis Ayres