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

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

Step 1 · Characteristic equations and first integrals

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dz}{2xy}. \tag{1} \] From the first two ratios: $y/x=c_1$. \tag{I} From the first and third, with $y=c_1x$: $dz=2c_1x\,dx$, so $z=c_1x^2+c_2$, i.e. \[ c_2 = z - xy. \tag{II} \] General solution: $z-xy=\Phi(y/x)$.

Step 2 · Test the initial condition

Parametrise $y=x$, $z=x^2$ by $t=x$. Then $c_1=y/x=1$ for every point, and $c_2=z-xy=t^2-t^2=0$ for every point. So the entire initial curve maps to the single point $(c_1,c_2)=(1,0)$ in parameter space — it lies along a single characteristic, and $\Phi$ is only constrained by $\Phi(1)=0$, not determined as a function.

Step 3 · Resolve the degeneracy

Indeed, the characteristic direction at a point of $y=x$ is $(P,Q,R)=(x,y,2xy)=(x,x,2x^2)$, which is exactly tangent to the curve $(t,t,t^2)$ (since $\tfrac{d}{dt}(t,t,t^2)=(1,1,2t)\propto(x,x,2x^2)$) — confirming the initial curve is itself a characteristic. Any $\Phi$ with $\Phi(1)=0$ gives a valid integral surface through the curve; taking the one-parameter family $\Phi(s)=\lambda(s-1)$: \[ z = xy + \lambda\,\frac{y-x}{x}, \] with the simplest choice $\lambda=0$ giving the particular solution $z=xy$.

Verification

At $y=x$: $z=x^2+\lambda\cdot0=x^2$ for every $\lambda$. $\checkmark$ With $p=y-\lambda y/x^2$, $q=x+\lambda/x$: \[ xp+yq = xy-\lambda y/x+xy+\lambda y/x = 2xy. \checkmark \]