Finding singular integral for a given pde

Question

The following pde is given and I need to find the complete integral and the singular integral if it exists:
$pqz=p^2(qx+p^2)+q^2(py+q^2)$
Now, I am able to find the complete integral as it is in Clairaut's form. So the complete integral is given by;
$z=ax+by+\frac{a^4+b^4}{ab}$
For finding the singular integral, I need to eliminate the arbitrary constants $a$ and $b$. So, I differentiate the above equation w.r.t. $a$ and w.r.t. $b$ , and I get;
$x+\frac{3a^2}{b}-\frac{b^3}{a^2}=0$
$y+\frac{3b^2}{a}-\frac{a^3}{b^2}=0$
Now, from here I am not able to proceed further and eliminate $a$ and $b$.
Any kind of help or hint will be of great help. Thanking everyone in advance!!

Answer 1

$x+\frac{3a^2}{b}-\frac{b^3}{a^2}=0\quad\implies\quad x=-\frac{3a^2}{b}+\frac{b^3}{a^2}$

$y+\frac{3b^2}{a}-\frac{a^3}{b^2}=0\quad\implies\quad y=-\frac{3b^2}{a}+\frac{a^3}{b^2}$

$z=ax+by+\frac{a^4+b^4}{ab}=a(-\frac{3a^2}{b}+\frac{b^3}{a^2})+b(-\frac{3b^2}{a}+\frac{a^3}{b^2})+\frac{a^4+b^4}{ab}$

After simplification : $\quad z= -\frac{a^4+b^4}{ab}$

The solution is obtained on parametric form : $$\begin{cases} x=-\frac{3a^2}{b}+\frac{b^3}{a^2} \\ y=-\frac{3b^2}{a}+\frac{a^3}{b^2} \\ z=-\frac{a^4+b^4}{ab} \end{cases}$$ There is no need to eliminate the parameters $a,b$ since the solution is already obtained.

Anyways eliminating $a,b$ would lead to an high degree polynomial equation. Analytical solving for $z(x,y)$ is not possible since the degree is higher than 4. The best way for practical use is numerical calculus starting from the above parametric form of solution.