Find the integral surface of yz p + xz q = xy, p = (∂ z)/(∂…

Find the integral surface of \[ yz\,p + xz\,q = xy, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $x=a$, $y^2+z^2=a^2$.

Step 1 · Characteristic equations

\[ \frac{dx}{yz} = \frac{dy}{xz} = \frac{dz}{xy}. \tag{1} \]

Step 2 · First integral

Multipliers $(x,y,-z)$: denominator $x\cdot yz+y\cdot xz-z\cdot xy = xyz$, numerator $x\,dx+y\,dy-z\,dz$. Equating to the first ratio (also denominator $yz$ matched via cross-multiplication) and simplifying: \[ y\,dy - z\,dz = 0 \;\Longrightarrow\; y^2-z^2=c_1. \tag{I} \]

Step 3 · Second integral

From the first two ratios directly: $xz\,dx=yz\,dy \Rightarrow x\,dx=y\,dy$. Integrating: \[ x^2-y^2 = c_2. \tag{II} \]

Step 4 · General solution

\[ x^2-y^2 = \Phi(y^2-z^2). \tag{$*$} \]

Step 5 · Apply the initial condition

At $x=a$, $y^2+z^2=a^2$, so $z^2=a^2-y^2$ and $y^2-z^2=2y^2-a^2$. With $t=2y^2-a^2$ (so $a^2-y^2=(a^2-t)/2$): $\Phi(t)=(a^2-t)/2$. Substituting $t=y^2-z^2$ into ($*$) and simplifying: \[ 2(x^2-y^2) = a^2-(y^2-z^2) \;\Longrightarrow\; 2x^2-y^2-z^2=a^2. \]

Verification

At $x=a$: $2a^2-y^2-z^2=a^2 \Rightarrow y^2+z^2=a^2$. $\checkmark$ Implicit differentiation of $2x^2-y^2-z^2=a^2$ gives $p=2x/z$, $q=-y/z$, so \[ yzp+xzq = 2xy - xy = xy. \checkmark \]