Example 1

Here the vector field is {-y, x}. The streamlines should look familiar.


  p1=PlotVectorField[{-y,x},{x,-5,5},{y,-5,5},Axes->True];


  DSolve[x'[t]/-y[t]==y'[t]/x[t],y[t],t]

Look at the solution. We need two branches to make the circle.

To get nice output (instead of a list of solutions, the format is:
Clear the variables.
DSolve[{x'[t]/f1==y'[t]/f2, initial conditions,y,t]
The line y1=Evaluate[y[t]/.Flatten[%][[1]]]/.x[t]->x;
is used to write y as a function x instead of x[t].


  Clear[x,y,y1,y2];
  DSolve[{x'[t]/-y[t]==y'[t]/x[t],y[0]==0,x[0]==2},y,t];
  y1=Evaluate[y[t]/.Flatten[%][[1]]]/.x[t]->x;
  DSolve[{x'[t]/-y[t]==y'[t]/x[t],y[0]==0,x[0]==2},y,t];
  y2=Evaluate[y[t]/.Flatten[%][[2]]]/.x[t]->x;
  y1
  y2


  p2=Plot[{y1,y2},{x,-3,3},AspectRatio->Automatic,
  PlotStyle->{RGBColor[1,0,0]}];


  Show[p1,p2];

Up to Streamlines