Question 2.29: Find the complete and singular solution (s), if any, of the ......

a) Putting

F(x, y, z, p, q)=1+p^2-z q

we obtain

\frac{\partial F}{\partial x} \frac{\partial F}{\partial x}+p \frac{\partial F}{\partial z}=-p q \text { and } \frac{\partial F}{\partial y}+q \frac{\partial F}{\partial z}=-q^2

Then from system 2.38 we get

\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)

\frac{-d p}{-p q}=\frac{-d q}{-q^2}

and thus we can take q = ap for an arbitrary constant a. Solving in p the quadratic equation 1+p^2-z \cdot(a p)=0, we obtain

p_{1,2}=\frac{-a z \pm \sqrt{a^2 z^2-4}}{2}          (2.46)

Let us take the sign ”+ ” in (2.46); then it holds

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

where b is another arbitrary constant, and thus the complete solution is

x+a y+b+\frac{z}{4} \sqrt{a^2 z^2-4}-\frac{1}{a} \ln \left|a z+\sqrt{a^2 z^2-4}\right|+\frac{a z^2}{4}=0 .

There is no singular solution.
Remark 2.29.1 The reader should check that any nonlinear first order PDE of the form F(p, q) = 0 has a first integral q = ap, for an arbitrary constant a.
Remark 2.29.2 If we took the sign ”-” in (2.46), we would get another complete solution. This shows that the complete solution of a nonlinear first order PDE is not unique; in fact, it even might have infinitely many complete solutions.
b) One easily gets the complete solution

z=a x+b y+a^2+a b+b^2 \quad(a, b \text { arbitrary constants }) .

Taking the partial derivatives in a and b of the complete solution, respectively, we obtain the system

x + 2a + b = 0, y + a + 2b = 0.

Solving it in a and b, we get the singular solution

z=\frac{1}{3}\left(x y-x^2-y^2\right)

Remark 2.29.3 The equation of the form z = px + qy + f(p, q), for some given function f is called “generalized Clairaut’s equation” .

c ) The fist integral is p=\frac{\sqrt{1-z^2}}{z \sqrt{1+a^2}}, and thus t he complete integral is

-\sqrt{1-z^2}=\frac{1}{\sqrt{1+a^2}}(x+a y)+b

The singular integrals are z = 1 and z = -1.