I would like to define a function based on the output of the following DSolve call:
DSolve[{b Varx^2 - 2 b Varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == V0x}, x[t], t]
I've tried to do the following:
x[t_] := DSolve[{b Varx^2 - 2 b Varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == V0x}, x[t], t]
It showed the message: $RecursionLimit::reclim: Recursion depth of 256 exceeded., so I believe I'm not doing it the right.
Any ideas?
You did not give numerical values for some of the parameters, so I made up some. For some, you might not get solutions, or get complex solution, so that is something you can look into since I do not know the physics of the problem.
Clear[x, t, b, varx, m, v0x]; y[t_] = x[t] /. First@DSolve[{b varx^2 - 2 b varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == v0x}, x[t], t] gives
(b t varx - m Log[m/(v0x - varx)] + m Log[b t - m/(-v0x + varx)])/b Then you can use the function y[t]
parms = {b -> 1, varx -> 2, m -> 1, x0 -> 1, v0x -> 0}; Plot[y[t] /. parms, {t, 0, 1}] 
D[y[t] /. parms, t] Out[48]= 2 + 1/(-(1/2) + t) etc...
Following Nasser's answer, here is a minor variation:
x[tt_, {b_, varx_, m_, x0_, v0x_}] := Module[{}, x[t_, {b, varx, m, x0, v0x}] = Block[{x, t}, x[t] /. First@ DSolve[{b varx^2 - 2 b varx x'[t] + b x'[t]^2 + m x''[t] == 0, x[0] == x0, x'[0] == v0x}, x[t], t] ]; x[tt, {b, varx, m, x0, v0x}] ] Then you can evaluate it by:
parms = {1, 2, 1, 1, 0}; x[4, parms] Plot[x[t, parms], {t, 0, 1}] etc. Note that ODE is computed just once for each parameter vector.
