Question 2.27: Assume equation (2.32) can be written in the form ∂z/∂x = f ......

Hint. Firstly take a = a( x, y) and b = b( x, y) in the complete solution

z = g(x,y,z,a,b).

Then, by the definition of the complete solution, replacing z, a and b with the functions g(x,y,z,a,b), \frac{\partial g}{\partial x}(x, y, z, a, b) \text { and } \frac{\partial g}{\partial y}(x, y, z, a, b) respectively, one gets an identity in (2.41) for arbitrary functions a and b.
Conversely, a function z = g(x, y, z, a, b) is a solution of (2.41) only if

\frac{\partial z}{\partial x}=\frac{\partial g}{\partial x} \text { and } \frac{\partial z}{\partial y}=\frac{\partial g}{\partial y} \text {. }

Show that this will hold only if

\begin{aligned}& \frac{\partial g}{\partial a}(x, y, a, b) \frac{\partial a}{\partial x}+\frac{\partial g}{\partial b}(x, y, a, b) \frac{\partial b}{\partial x}=0 \\& \frac{\partial g}{\partial a}(x, y, a, b) \frac{\partial a}{\partial y}+\frac{\partial g}{\partial b}(x, y, a, b) \frac{\partial b}{\partial y}=0 .\end{aligned}         (2.42)

Putting \frac{\partial g}{\partial a}(x, y, a, b)=0 \text { and } \frac{\partial g}{\partial b}(x, y, a, b)=0, and, if possible, eliminating a and b from equations (2.42), we obtain the singular solution. Note that it depends neither from arbitrary constants nor from arbitrary functions.
Next, assume it holds \frac{\partial(a, b)}{\partial(x, y)}=0, where

\frac{\partial(a, b)}{\partial(x, y)}=\left|\begin{array}{ll}\frac{\partial a}{\partial x} & \frac{\partial a}{\partial y} \\\frac{\partial b}{\partial x} & \frac{\partial b}{\partial y}\end{array}\right|,

and let a and b be some non constant functions. Then the functions a and bare not functionally independent, i.e., there exists a differentiable function w such that b = w(a). This leads to the general solution of (2.41).
Finally, check that any solution of (2.41) reduces to one of the following three ones: singular, complete or general.