T = ½p (x12 + x22) + ½p′x32 + q (x1y1 + x2y2) + q′x3y3 + ½r (y12 + y22) + ½r′y32

(1)

so that a fourth integral is given by

dy3 / dt = 0, y3 = constant;

(2)

dx3= x1 (qx2 + ry2) − x2 (qx1 + ry1) = r (x1y2 − x2y1),
dt

(3)

1( dx3) 2= (x12 + x22) (y12 + y22) − (x1y1 + x2y2)2
r2 dt
= (x12 + x22) (y12 + y22) − (FG − x3y3)2 = (x12 + x22) (y12 + y22 + y32 − G2) − (Gx3 − Fy3)2,

(4)