Direction Fields.

We will use the command PlotVectorField to plot direction fields.  The equation y' = f(t,y) tells us that at the point (t,y) we want to plot a short line with slope f(t,y).  The vector (1,f(t,y)) has slope f(t,y), so we simply plot the vector field (1,f(t,y)), (or i + f(t,y)j),  but we add an option ScaleFunction->(1&) to make all the arrows the same length.  (1&) is a pure function in Mathematica standing for the constant function with value 1.  We need a Graphics package, or Add-On, to plot a vector field..   You should expect plotting of vector fields to be slow.

Needs["Graphics`PlotField`"]

Example 1.  y' =  t - y

PlotVectorField[{1, t - y}, {t, -3, 5}, {y, -3, 3}, 
      Axes -> True, ScaleFunction -> (1&), Ticks -> None, 
      Frame -> True, AspectRatio -> 1];

It appears that all solutions are asymptotic to the line y=t-1.Indeed, using an integrating factor
to solve the differential equation, one finds that the general solution is   y = (t-1) + C exp(-t).
We plot this solution for C = {-3,-2, -1, 0, 1, 2, 3}:

\(Plot[
      Evaluate[Table\ [\ 
          \((t - 1)\)\  + \ C\ \ Exp[\(-t\)]\ , {C, \(-3\), 3, 1}]], 

      		{t, \(-3\), 5}, \ AspectRatio -> 1, \ PlotRange\  -> \ {\(-3\), 3}]
      \)

Caution:  If you execute a PlotVectorField command before you do the Needs command, either restart or execute the following commands.

Remove[PlotVectorField, ScaleFunction]Needs["Graphics`PlotField`"]

A palette for direction fields.
This button was made by choosing "Create Table/Matrix/Palette..." from the Input menu, and then filling in the command we want.  The squares are given by pressing the key marked "esc" followed by the letters spl followed by the esc key again..    Next select the cell and then "Generate Palette from Selection" from the File menu.  Palettes like this are especially useful when  there a re many options to type into a command.

({{\(PlotVectorField[{1, \[SelectionPlaceholder]}, {t, 
                \[SelectionPlaceholder], \[SelectionPlaceholder]}, {y, 
                \[SelectionPlaceholder], \[SelectionPlaceholder]}
	, 
              PlotPoints -> 15, Axes -> True
	, 
              ScaleFunction -> \((1&)\), Ticks -> None
	, 
              Frame -> True, AspectRatio -> 1]; \)}, {\[Placeholder]}})

Example 2.  y' =  y + 2
Click on the palette button and fill in the blanks.  If you want more vectors, raise the PlotPoints option.

PlotVectorField[{1, t (E^(-2 t)) - 2 y}, {t, -2, 3}, {y, 
        -6, 6}, PlotPoints -> 15, Axes -> True, 
      ScaleFunction -> (1&), Ticks -> None, Frame -> True, 
      AspectRatio -> 1];

My guess is that all solutions tend to 0 as t tends to infinity.

PlotVectorField[{1, (t/y)}, {t, -4, 4}, {y, -3, 3}, 
      PlotPoints -> 20, Axes -> True, 
      ScaleFunction -> (1&), Ticks -> None, Frame -> True, 
      AspectRatio -> 1];

Some solutions approach the negative y-axis.  Others approach the line y = t or the line y = - t.