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

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

Step 1 · Characteristic equations and first integral

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

Step 2 · Second integral

Multipliers $(x,-y,-1)$ give combined denominator $x^2+2y^2-(x^2-y^2)=3y^2$ and numerator $x\,dx-y\,dy-dz$. Equating to the second ratio and simplifying (dividing by $y$): \[ 2\,dz = 2x\,dx + y\,dy \;\Longrightarrow\; 2z-x^2-\frac{y^2}{2}=c_2. \tag{II} \]

Step 3 · General solution

\[ z = \frac{x^2}{2}+\frac{y^2}{4}+\frac{1}{2}\Phi(x^2y). \tag{$*$} \]

Step 4 · Apply the initial condition

Set $y=1$, $z=x^2$: $x^2=\tfrac{x^2}{2}+\tfrac14+\tfrac12\Phi(x^2)$, so $\Phi(t)=t-\tfrac12$ with $t=x^2$. Substituting $t=x^2y$ into ($*$): \[ z = \frac{x^2}{2}+\frac{y^2}{4}+\frac{x^2y}{2}-\frac{1}{4} = \frac{x^2(1+y)}{2}+\frac{y^2-1}{4}. \]

Verification

At $y=1$: $z=x^2$. $\checkmark$ Compute $p=x(1+y)$, $q=\tfrac{x^2}{2}+\tfrac{y}{2}$: \begin{align*} xp-2yq &= x^2(1+y) - yx^2-y^2 = x^2+x^2y-x^2y-y^2 = x^2-y^2. \checkmark \end{align*}