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

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

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{x} = \frac{dy}{xz+y} = \frac{dz}{z}. \tag{1} \] From the first and third ratios: $z/x=c_1$. \tag{I}

Step 2 · Second integral

From $(xz+y)\,dx=x\,dy$: $x\,dy-y\,dx=xz\,dx$. With $z=c_1x$ along the characteristic, the right side is $c_1x^2\,dx$; dividing by $x^2$: \[ d\!\left(\frac{y}{x}\right) = c_1\,dx \;\Longrightarrow\; \frac{y}{x}=c_1x+c_2. \] Hence \[ c_2 = \frac{y}{x} - c_1x = \frac{y-xz}{x}. \tag{II} \]

Step 3 · General solution

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

Step 4 · Apply the initial condition

Solving $x-y=1$, $x+y=2z$ gives $y=x-1$, $z=x-\tfrac12$. Parametrise by $t=x$: $x=t$, $y=t-1$, $z=t-\tfrac12$. On this curve, \[ c_1 = \frac{z}{x} = 1-\frac{1}{2t}, \qquad c_2 = \frac{y-xz}{x} = -t+\frac32-\frac1t. \] Eliminating $t$ (via $t=1/[2(1-c_1)]$) and substituting back gives, after simplification, the implicit integral surface \[ 2xy - 2yz - 2x^2z + 2xz^2 + 2x^2 - 5xz + 4z^2 = 0. \]

Verification

Substituting $y=x-1$, $z=x-\tfrac12$ into the boxed relation, every power of $x$ cancels identically, confirming the initial curve lies on the surface. $\checkmark$
Checking $c_1=z/x$ and $c_2=(y-xz)/x=y/x-z$ against $P(c)_x+Q(c)_y+R(c)_z=0$ with $P=x$, $Q=xz+y$, $R=z$: \[ x\!\left(-\frac{z}{x^2}\right)+(xz+y)\cdot0+z\cdot\frac1x=0, \qquad x\!\left(-\frac{y}{x^2}\right)+(xz+y)\cdot\frac1x+z\cdot(-1)=0. \checkmark \]