Find the integral surface of (x-z) p + (y-z) q = 2z, p = (∂…

Find the integral surface of \[ (x-z)\,p + (y-z)\,q = 2z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $x-y=2$, $z+2x=1$.

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{x-z} = \frac{dy}{y-z} = \frac{dz}{2z}. \tag{1} \] Multipliers $(1,-1,0)$: $\dfrac{d(x-y)}{x-y}=\dfrac{dz}{2z}$. Integrating: \[ \frac{(x-y)^2}{z} = c_1. \tag{I} \]

Step 2 · Second integral

From $2z\,dx=(x-z)\,dz$: $\dfrac{dx}{dz}-\dfrac{x}{2z}=-\dfrac12$. With integrating factor $z^{-1/2}$: $\dfrac{d(xz^{-1/2})}{dz}=-\tfrac12z^{-1/2}$, so $xz^{-1/2}=-z^{1/2}+c_2$, giving \[ c_2 = \frac{x+z}{\sqrt{z}}. \tag{II} \]

Step 3 · General solution

\[ \frac{x+z}{\sqrt z} = \Phi\!\left(\frac{(x-y)^2}{z}\right). \tag{$*$} \]

Step 4 · Apply the initial condition

Parametrise by $t=x$: $y=t-2$, $z=1-2t$. Then $c_1=4/(1-2t)$, $c_2=(1-t)/\sqrt{1-2t}$. Expressing $t$ (hence $c_2$) in terms of $c_1$ gives $c_2=(c_1+4)/(4\sqrt{c_1})$. Substituting $c_1=(x-y)^2/z$ into ($*$) and clearing denominators: \[ 4(x-y)(x+z) = (x-y)^2+4z, \] which factors as \[ (3x+y)(x-y) + 4z(x-y-1) = 0. \]

Verification

Substituting $y=x-2$, $z=1-2x$ into the boxed relation cancels all powers of $x$ identically. $\checkmark$ Implicit differentiation gives $p=-\tfrac{6x-2y+4z}{4(x-y-1)}$, $q=\tfrac{x+y+2z}{2(x-y-1)}$; substituting into $(x-z)p+(y-z)q$ and using the surface relation $(3x+y)(x-y)=-4z(x-y-1)$ yields $2z$. $\checkmark$