yzp²-p=0 to be solved using Char-pit’s method.

F(x,y,z,p,q)=0 where p=∂z/∂x and q=∂z/∂y where z=f(x,y).

By definition, F=yzp²-p=0.

dx/(∂F/∂p)=dx/(2pyz-1)=

dy/(∂F/∂q)=dy/0= (implies dy=0)

dz/(p∂F/∂p+q∂F/∂q)=dz/(p(2pyz-1))=

dp/-(∂F/∂x+p∂F/∂z)=dp/(-p³y)=-dp/(p³y)=

dq/-(∂F/∂y+q∂F/∂z)=dq/-(zp²+yp²q)=-dq/(p²(z+yq)).

Since dx/(2pyz-1)=dz/(p(2pyz-1)), dx=dz/p, dz=pdx⇒z=f(x) (no y terms) and p=dz/dx.

Also, dp/(py)=dq/(z+yq)=dq/z, since q=∂z/∂y=0.

In normal differentiation, for z=f(x,y), dz/dx=∂z/∂x+∂z/∂y.dy/dx, so dz=pdx+qdy. It follows that if pdx=dz, qdy=0 and dy/dx=0.

F=yz(dz/dx)²-dz/dx=0; yzdz/dx=1 assuming dz/dx≠0⇒z≠a, where a is constant.

dy/dx=(∂F/∂q)/(∂F/∂p)=0, because ∂F/∂q=0, so y=c, where c is constant.

Therefore czdz=dx, and, integrating wrt x: cz²/2=x+b, where b is constant.

So z=√(2(x+b)/c) making f(x)=√(2(x+b)/c) and y=c.