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

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

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{xy} = \frac{dy}{-y^2} = \frac{dz}{x}. \tag{1} \] From the first two ratios: $dx/x=-dy/y$. Integrating: $xy=c_1$. \tag{I}

Step 2 · Second integral

From $x\,dx=xy\,dz$ with $y=c_1/x$: $dz=x\,dx/c_1$. Integrating: $z=x^2/(2c_1)+c_2 = x/(2y)+c_2$, so \[ c_2 = z - \frac{x}{2y}. \tag{II} \]

Step 3 · General solution

\[ z = \frac{x}{2y} + \Phi(xy). \tag{$*$} \]

Step 4 · Apply the initial condition

At $x=a$, $z=(a^2+2)/(2ay)$: $\dfrac{a^2+2}{2ay} = \dfrac{a}{2y}+\Phi(ay)$, so $\Phi(ay)=\dfrac{1}{ay}$. With $t=ay$: $\Phi(t)=1/t$. Substituting $t=xy$: \[ z = \frac{x}{2y} + \frac{1}{xy}. \]

Verification

At $x=a$: $z=\dfrac{a}{2y}+\dfrac{1}{ay}=\dfrac{a^2+2}{2ay}$, i.e. $2ayz=a^2+2$. $\checkmark$ Compute $p=\tfrac1{2y}-\tfrac1{x^2y}$, $q=-\tfrac{x}{2y^2}-\tfrac1{xy^2}$: \[ xyp-y^2q = \frac{x}{2}-\frac1x+\frac{x}{2}+\frac1x = x. \checkmark \]