Example 2

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


  p3=PlotVectorField[{y,x},{x,-5,5},{y,-5,5},Axes->True];


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

Note from the solution: there are two branches of the hyperbola.


  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


  p4=Plot[{y1,y2},{x,-5,5},AspectRatio->Automatic,
  PlotStyle->{RGBColor[1,0,0]}];


  Show[p3,p4];

Up to Streamlines