Find the integral surface of xy^3 p + x^2z^2 q = y^3z, p =…

Find the integral surface of \[ xy^3\,p + x^2z^2\,q = y^3z, \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}, \] that passes through the curve $x=-z^3$, $y=z^2$.

Step 1 · Characteristic equations and first integral

\[ \frac{dx}{xy^3} = \frac{dy}{x^2z^2} = \frac{dz}{y^3z}. \tag{1} \] From the first and third ratios, $y^3$ cancels: $dx/x=dz/z \Rightarrow z/x=c_1$. \tag{I}

Step 2 · Second integral

From $x^2z^2\,dx=xy^3\,dy$: $xz^2\,dx=y^3\,dy$. Using $z=c_1x$: $c_1^2x^3\,dx=y^3\,dy$. Integrating: $c_1^2x^4/4=y^4/4+\text{const}$, and substituting $c_1=z/x$: \[ x^2z^2 - y^4 = c_2. \tag{II} \]

Step 3 · General solution

\[ x^2z^2 - y^4 = \Phi\!\left(\frac{z}{x}\right). \tag{$*$} \]

Step 4 · Apply the initial condition

Parametrise $x=-z^3$, $y=z^2$ by $t=z$: $x=-t^3$, $y=t^2$, $z=t$. Then $c_1=-1/t^2$ and $c_2=x^2z^2-y^4=t^8-t^8=0$ for every $t$, so $\Phi\equiv0$. Hence $x^2z^2-y^4=0$.

Verification

On the curve: $x^2z^2=t^8=y^4$. $\checkmark$ From $x^2z^2=y^4$: $p=-z/x$, $q=2y^3/(x^2z)$, so \[ xy^3p+x^2z^2q = -y^3z+2y^3z = y^3z. \checkmark \]