Find the general solution of xz (∂ z)/(∂ x) + yz (∂ z)/(∂ y)…

Find the general solution of \[ xz\,\frac{\partial z}{\partial x} + yz\,\frac{\partial z}{\partial y} = -xy, \] and then find the integral surface passing through the curve $y = x^2$, $z = x^3$.

Step 1 · Reduce to a linear PDE

With $w=z^2$, $\partial_x w = 2z\,\partial_x z$, $\partial_y w = 2z\,\partial_y z$. Multiplying the original equation by $2$: \[ x\,\frac{\partial w}{\partial x} + y\,\frac{\partial w}{\partial y} = -2xy. \tag{1} \]

Step 2 · Characteristic equations

\[ \frac{dx}{x} = \frac{dy}{y} = \frac{dw}{-2xy}. \tag{2} \] First integral from $\dfrac{dx}{x}=\dfrac{dy}{y}$: $\dfrac{y}{x}=c_1$. (I)
Second integral: along characteristics $y=c_1x$, so $\dfrac{dx}{x}=\dfrac{dw}{-2c_1x^2}$ gives $dw=-2c_1x\,dx$, hence $w=-c_1x^2+c_2$. Substituting $c_1=y/x$: \[ c_2 = w + xy = z^2 + xy. \tag{II} \]

Step 3 · General solution

The general solution is $\Phi(c_1,c_2)=0$, i.e. \[ z^2 + xy = \Phi\!\left(\frac{y}{x}\right), \] where $\Phi$ is an arbitrary differentiable function.

Step 4 · Integral surface through $y=x^2$, $z=x^3$

Parameterise the curve by $x=t$: $x=t$, $y=t^2$, $z=t^3$. On this curve, \[ c_1 = \frac{y}{x} = t, \qquad c_2 = z^2 + xy = t^6 + t^3, \] so $\Phi(u) = u^6+u^3$ with $u=y/x$. Substituting back: \[ z^2 + xy = \left(\frac{y}{x}\right)^6 + \left(\frac{y}{x}\right)^3. \]

Verification

Curve on the surface: substituting $y=x^2$, $z=x^3$ into $y^6/x^6+y^3/x^3-xy$ gives $x^6+x^3-x^3=x^6=z^2$. $\checkmark$
Surface satisfies the PDE: differentiating $z^2+xy=\Phi(y/x)$ with respect to $x$ and $y$ and eliminating $\Phi'(y/x)$ between the two relations yields, after simplification, \[ xzp + yzq = -xy. \checkmark \]