Example 2

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

In[7]:=

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

In[8]:=

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

Out[8]=

                             2
  {{y[t] -> -Sqrt[C[1] + x[t] ]}, 
   
                             2
    {y[t] -> Sqrt[C[1] + x[t] ]}}

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

In[9]:=

  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

Out[9]=

              2
  -Sqrt[-4 + x ]

Out[10]=

             2
  Sqrt[-4 + x ]

In[11]:=

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

In[12]:=

  Show[p3,p4];

Up to Streamlines