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

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

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{y+2z^2} = \frac{dy}{-2x^2z} = \frac{dz}{x^2}. \tag{1} \] From the second and third ratios: $x^2\,dy=-2x^2z\,dz \Rightarrow dy=-2z\,dz$. Integrating: \[ y+z^2 = c_1. \tag{I} \]

Step 2 · Second integral

Along a characteristic $y+2z^2=c_1+z^2$; substituting into the first ratio: \[ x^2\,dx = (c_1+z^2)\,dz. \] Integrating: $x^3/3 = c_1z+z^3/3+c_2$; substituting $c_1=y+z^2$ and simplifying: \[ c_2 = \frac{x^3-4z^3}{3} - yz. \tag{II} \]

Step 3 · General solution

\[ \frac{x^3-4z^3}{3} - yz = \Phi(y+z^2). \tag{$*$} \]

Step 4 · Apply the initial condition

Parametrise $x=z$, $y=x^2$ by $t=x$: $x=t$, $y=t^2$, $z=t$. On this curve $c_1=2t^2$ and $c_2=-2t^3$, so with $t^2=c_1/2$: $c_2=-2(c_1/2)^{3/2}=-\sqrt{c_1^3/2}$, giving $\Phi(s)=-\sqrt{s^3/2}$. Substituting $s=y+z^2$ and squaring: \[ 2\!\left(\frac{x^3-4z^3}{3}-yz\right)^{\!2} = (y+z^2)^3. \]

Verification

On $x=z=t$, $y=t^2$: left side $=2(-2t^3)^2=8t^6$, right side $=(2t^2)^3=8t^6$. $\checkmark$ Since $c_1=y+z^2$ and $c_2=(x^3-4z^3)/3-yz$ both satisfy $P(c)_x+Q(c)_y+R(c)_z=0$ (with $P=y+2z^2$, $Q=-2x^2z$, $R=x^2$), every surface $c_2=\Phi(c_1)$ solves the PDE. $\checkmark$