Plotting solution curves of solutions defined implicitly.

When the solutions of a first order differential equation are defined implicitly, as in the case with separable equations, the command ContourPlot can be used to graph the solutions.

?ContourPlot

Options[ContourPlot]

The meaning and usage of the options can as usual be found by selecting the command ContourPlot and choosing "Find in Help..." from the Help menu.  There we find, for example, that the option Contours->{a,b,c,...} will graph the contours with specific values of the constant.
Suppose the solution is given in the form g(x,y) = c.  We can then use the following model, with suitable editing for each example.

g[x_,y_]=?

curves=ContourPlot[g[x,y],{x,-3,3},{y,-3,3},
					Axes->True,Contours->15,
					ContourShading->False,
					ContourSmoothing->True,
					PlotPoints->25]

Example 1.  The equation y' = (x^2)/(1 - (y^2)) on page 34 of the text has solution (x^3)+ 3y - (y^3) = c.

g[x_, y_] = (x^3) - 3  y + (y^3)

curves=ContourPlot[g[x,y],{x,-3,3},{y,-3,3},
					Axes->True,Contours->25,
					ContourShading->False,
					ContourSmoothing->True,
					PlotPoints->25]

Note that the denominator of  the right hand side of the equation f(t, y) is zero when y is 1 or -1, and that all the solution curves have a vertical when y is 1 or -1.

If we wanted the solution curve for which y = 2 when x = 1 , we set c = g(1,2)

g[x_, y_] = (x^3) - 3  y + (y^3)

curves=ContourPlot[g[x,y],{x,-3,3},{y,-3,3},
					Axes->True,Contours->{g[1,2]},
					ContourShading->False,
					ContourSmoothing->True,
					PlotPoints->25]

The domain of the solution with y = 2 when x = 1 can be roughly found by going to the curve at the point (1,2) and seeing how far you can go to the left and right and still have a well-defined function of x, in this case from about -2.5 to about 1.35.

Example 2.  The equation y' = (y cos x)/(1+2(y^2)) on page 36 has solution
 ln |y| +(y^2) = sin x + C.

g[x_, y_] = Log[Abs[y]] + (y^2) - Sin[x]

curves=ContourPlot[g[x,y],{x,-3,3},{y,.1,3},
					Axes->True,Contours->15,
					ContourShading->False,
					ContourSmoothing->True,
					PlotPoints->25]

The solution which has y=1 when x=0 is given by

g[x_, y_] = Log[Abs[y]] + (y^2) - Sin[x] - 1

curves=ContourPlot[g[x,y],{x,-3,3},{y,.1,3},
					Axes->True,Contours->{g[0,1]},
					ContourShading->False,
					ContourSmoothing->True,
					PlotPoints->25]