(* Solution curves of x'[t] == x[t]^2 + t^2 - 1 for
   given initial values. Background is shaded to show
   slope of curves; black circle corresponds to zero
   slope (nullcline). This example is on p. 26 of 
   "VisualDSolve" by Dan Schwalbe and Stan Wagon  *)

Needs["VisualDSolve`"];

VisualDSolve[x'[t] == x[t]^2 + t^2 - 1,
    {t, -2.5, 2.5}, {x, -2.5, 2.5},
    InitialValues ->
      Table[1.5{Cos[t], Sin[t]}, {t, 0, 2 Pi, Pi/10}],
    ShowInitialValues -> True, StayInWindow -> True,
    IsoclineShading -> True,
    IsoclineStyle -> AbsoluteThickness[2],
    Rainbow -> True, PlotStyle -> AbsoluteThickness[2],
    Contours -> 30, Isoclines -> True,
    IsoclinePlotPoints -> 100,
    AspectRatio -> 1];