Find the general solution of (xy^3 - 2x^4)p + (2y^4 - x^3…

Find the general solution of \[ \left(xy^3 - 2x^4\right)p + \left(2y^4 - x^3 y\right)q = 9z\!\left(y^3 - x^3\right), \qquad p = \frac{\partial z}{\partial x},\quad q = \frac{\partial z}{\partial y}. \]

Step 1 · Characteristic equations

\[ \frac{dx}{x(y^3-2x^3)} = \frac{dy}{y(2y^3-x^3)} = \frac{dz}{9z(y^3-x^3)}. \tag{1} \]

Step 2 · First integral

Multipliers $(1/x,1/y,0)$ give \[ \frac{dx/x+dy/y}{(y^3-2x^3)+(2y^3-x^3)} = \frac{d\ln(xy)}{3(y^3-x^3)} = \frac{dz}{9z(y^3-x^3)}. \] The factor $(y^3-x^3)$ cancels: $dz/z=3\,d\ln(xy)$. Integrating: \[ \frac{z}{(xy)^3} = c_1. \tag{I} \]

Step 3 · Second integral

From the first two ratios: $y(2y^3-x^3)\,dx=x(y^3-2x^3)\,dy$. Substitute $u=x^3$, $w=y^3$ ($dx=du/(3x^2)$, $dy=dw/(3y^2)$): \[ w(2w-u)\,du = u(w-2u)\,dw. \] With $s=u/w$ ($u=sw$, $du=s\,dw+w\,ds$): \[ (2-s)w\,ds = -s(1+s)\,dw \;\Longrightarrow\; \frac{(2-s)\,ds}{s(1+s)} = -\frac{dw}{w}. \] Partial fractions $\dfrac{2-s}{s(1+s)}=\dfrac{2}{s}-\dfrac{3}{1+s}$ give, on integrating, $\dfrac{s^2}{(1+s)^3}\,w=c$, i.e. $\dfrac{u^2w}{(u+w)^3}=c$. Substituting back $u=x^3$, $w=y^3$: \[ \frac{x^6y^3}{(x^3+y^3)^3} = c^3 \;\Longrightarrow\; \frac{x^2y^2}{x^3+y^3} = c_2. \tag{II} \]

Step 4 · General solution

The general solution $c_1=\Phi(c_2)$ is: \[ \frac{z}{x^3 y^3} = \Phi\!\left(\frac{x^2 y^2}{x^3+y^3}\right), \] where $\Phi$ is an arbitrary differentiable function.

Verification

First integral: with $c_1=z/(x^3y^3)$, $c_{1x}=-3z/(x^4y^3)$, $c_{1y}=-3z/(x^3y^4)$, $c_{1z}=1/(xy)^3$: \[ Pc_{1x}+Qc_{1y}+Rc_{1z} = \frac{z}{x^3y^3}\!\left[-3(y^3-2x^3)-3(2y^3-x^3)+9(y^3-x^3)\right]=0. \checkmark \] Second integral: with $c_2=x^2y^2/(x^3+y^3)$, $c_{2x}=\dfrac{xy^2(2y^3-x^3)}{(x^3+y^3)^2}$, $c_{2y}=\dfrac{x^2y(2x^3-y^3)}{(x^3+y^3)^2}$: \[ Pc_{2x}+Qc_{2y} = \frac{x^2y^2(2y^3-x^3)}{(x^3+y^3)^2}\!\left[(y^3-2x^3)+(2x^3-y^3)\right] = 0. \checkmark \] Since both first integrals satisfy the characteristic condition, the boxed relation solves the PDE for every choice of $\Phi$. $\checkmark$