Given the partial differential equation (z-y) p - (z+x) q =…
Given the partial differential equation
\[ (z-y)\,p - (z+x)\,q = x+y. \]
1° Find the general integral.
2° Find the general form of quadric surfaces satisfying the equation.
3° Find the Cauchy integral with $y=x+1$, $z=0$.
4° Investigate the surface from 3° and reduce it to canonical form.
Explanation of Method
Both multiplier combinations $(1,1,1)$ and $(x,y,z)$ make the combined denominator collapse nicely, giving both first integrals of the Lagrange PDE directly; the quadric and Cauchy sub-problems then follow by substitution, and the final quadric is classified by completing the square.
Step 1 · Part 1: General integral
\[ \frac{dx}{z-y}=\frac{dy}{-(z+x)}=\frac{dz}{x+y}. \] Multipliers $(1,1,1)$: denominator $(z-y)-(z+x)+(x+y)=0$, so $dx+dy+dz=0$ identically: \[ x+y+z=c_1. \tag{I} \] Multipliers $(x,y,z)$: denominator $x(z-y)-y(z+x)+z(x+y)=2x(z-y)$; equating to the first ratio gives $x\,dx+y\,dy+z\,dz=2x\,dx$, i.e. $y\,dy+z\,dz=x\,dx$: \[ x^2-y^2-z^2 = c_2. \tag{II} \] General integral: \[ x^2-y^2-z^2 = \Phi(x+y+z), \] where $\Phi$ is an arbitrary differentiable function.
Step 2 · Part 2: Quadric surfaces
Taking $\Phi(s)=As^2+Bs+C$ (the general quadratic) keeps the equation a second-degree polynomial in $x,y,z$: \[ x^2-y^2-z^2 = A(x+y+z)^2+B(x+y+z)+C, \] for arbitrary constants $A,B,C$.
Step 3 · Part 3: Cauchy integral through $y=x+1$, $z=0$
Parametrise by $t=x$: $x=t$, $y=t+1$, $z=0$. Then $c_1=2t+1$ and $c_2=t^2-(t+1)^2=-2t-1=-c_1$, so $\Phi(s)=-s$: \[ x^2-y^2-z^2 = -(x+y+z) \;\Longrightarrow\; x^2-y^2-z^2+x+y+z=0. \]
Step 4 · Part 4: Canonical form
Completing the square with $X=x+\tfrac12$, $Y=y-\tfrac12$, $Z=z-\tfrac12$: \[ X^2-Y^2-Z^2 = -\tfrac14 \;\Longrightarrow\; Y^2+Z^2-X^2=\tfrac14, \] i.e. $\dfrac{Y^2}{(1/2)^2}+\dfrac{Z^2}{(1/2)^2}-\dfrac{X^2}{(1/2)^2}=1$: a one-sheeted hyperboloid of revolution with semi-axes $\tfrac12$, axis parallel to $Ox$, centred at $\left(-\tfrac12,\tfrac12,\tfrac12\right)$.
Verification
First integrals: with $P=z-y$, $Q=-(z+x)$, $R=x+y$, both
$c_1=x+y+z$ and $c_2=x^2-y^2-z^2$ satisfy $Pc_x+Qc_y+Rc_z=0$ directly by the
multiplier constructions above. $\checkmark$
Cauchy surface: at $y=x+1$, $z=0$: $x^2-(x+1)^2+x+(x+1)=0$. $\checkmark$
For $F=x^2-y^2-z^2+x+y+z$, the condition $PF_x+QF_y+RF_z=0$ expands to
\[ (z-y)(2x+1)-(z+x)(-2y+1)+(x+y)(-2z+1) = 0 \]
after full expansion (all twelve terms cancel in pairs), confirming $F=0$ solves
the PDE. $\checkmark$
Result
The general integral, quadric family, and Cauchy surface (in canonical form) are: