Question 2.39: Solve the PDE p² + q² + pq - qx - py - 2z + xy = o....

I mode. From the system (2.38) we have

\frac{d x}{\frac{\partial F}{\partial p}}=\frac{d y}{\frac{\partial F}{\partial q}}=\frac{d z}{p \frac{\partial F}{\partial p}+q \frac{\partial F}{\partial q}}=\frac{-d p}{\frac{\partial F}{\partial x}+p \frac{\partial F}{\partial z}}=\frac{-d q}{\frac{\partial F}{\partial y}+q \frac{\partial F}{\partial z}}          (2.38)

Now we easily find one first integral, namely

x+a=p, \quad \text { and thus } \Psi_1=p-x \text {. }

Another first integral is

\Psi_2=q-y

Since these two first integrals are in involution (check that!), it follows that the complete solution is

z=\frac{1}{2}\left((x+a)^2+(y+b)^2+a b\right) .

II mode. As in mode I, we find that p = x + a is a first integral. From the given equation it follows

q=-\frac{a}{2}+\sqrt{2 z+a y-(x+a)^2+\frac{a^2}{4}} .

Thus we obtain Pfaff’s equation

d z=(x+a) d x+\left(-\frac{a}{2}+\sqrt{2 z+a y-(x+a)^2+\frac{a^2}{4}}\right) d y

or

\frac{d z+\frac{a}{2} d y-(x+a) d x}{\sqrt{2 z+a y-(x+a)^2+\frac{a^2}{4}}}=0 .

Hence the complete solution is

y+b+\frac{a}{2}=\sqrt{2 z+a y-(x+a)^2+\frac{a^2}{4}}