Find the integral surface of x p + z q = y, p = (∂ z)/(∂ x)…
Find the integral surface of \[ x\,p + z\,q = y, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $y=2z$, $x+2y=z$.
Explanation of Method
The first integral is immediate from the symmetric last two ratios; the second uses the sum combination $(1,1)$ on $y,z$. On the given initial curve the second invariant turns out to be constant, forcing $\Phi$ to be a constant function.
Step 1 · Characteristic equations and first integral
\[ \frac{dx}{x} = \frac{dy}{z} = \frac{dz}{y}. \tag{1} \] From the last two ratios: $y\,dy=z\,dz \Rightarrow y^2-z^2=c_1$. \tag{I}
Step 2 · Second integral
Summing the last two ratios: $\dfrac{d(y+z)}{y+z}=\dfrac{dx}{x}$. Integrating: \[ \frac{y+z}{x} = c_2. \tag{II} \]
Step 3 · General solution
\[ \frac{y+z}{x} = \Phi(y^2-z^2). \tag{$*$} \]
Step 4 · Apply the initial condition
From $y=2z$ and $x+2y=z$: $z=y/2$, $x=z-2y=-\tfrac{3y}{2}$. Parametrise by $t=y$: $x=-\tfrac{3t}{2}$, $y=t$, $z=\tfrac{t}{2}$. Then $c_2=(y+z)/x = (3t/2)/(-3t/2)=-1$ for every $t$, so $\Phi$ is the constant function $\Phi\equiv-1$. Substituting into ($*$): \[ \frac{y+z}{x} = -1 \;\Longrightarrow\; x+y+z=0. \]
Verification
On the initial curve, $x+3z=-3z+3z=0$ (using $x=-3z$). $\checkmark$ From $x+y+z=0$: $p=q=-1$, and $xp+zq=-(x+z)=y$ (since $x+z=-y$). $\checkmark$
Result
The integral surface is: