<s>[INST] How to solve this pde equation: $(p^2 + q^2)y = qz$ My attempt at solution
$p^2 y = qz - q^2 = a... (I)$
This equation is of the form $f_1(x,p) = f_2(y,q)$.
Its solution is given by $dz=pdx + qdy$, upon integrating this we get value of $z$.
From  (I)
$-yq_2 + zq - a = 0$, solving the quadratic equation for $q$, we get
$q=\frac{-z±\sqrt{z^2-4ay}}{-2y}$.
Taking the positive value only, $q=\frac{-z+\sqrt{z^2-4ay}}{-2y}$ .
Also, from (I), $p^2 y = a$, therefore $p=\frac{\sqrt{a}}{y}$.
Therfore $dz=\frac{\sqrt{a}{y}} dx + \frac{-z+\sqrt{z^2-4ay}}{-2y}dy$ .
I can't get any further. I know, how to solve by charpit's method, but my book mentions that I solve it without using charpit.
The answer is given: $z^2=(cx+a)^2 + c^2 y^2$ .
 [/INST] $let\ f(x,y,z,p,q)=(p^2+q^2)y-qz=0 \hspace{1cm} \to (1)$
Now,
$f_x=0$
$f_y=p^2+q^2$
$f_z=-q$
$f_p=2py$
$f_q=2qy-z$
The auxillary equations of the given DE are
$\dfrac{dp}{f_x+pf_z}=\dfrac{dq}{f_y+q.f_z}=\dfrac{dz}{-p.f_p-q.f_q}=\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}$
$\dfrac{dp}{-pq}=\dfrac{dq}{p^2+q^2-q^2}=\dfrac{dz}{-p(2py)-q(2qy-z)}=\dfrac{dx}{-2py}=\dfrac{dy}{z-2qy}\hspace{1cm} \to (2)$
From (2),
$\dfrac{dp}{-pq}=\dfrac{dq}{p^2}$
$-pdp=qdq $
On Integration,
$ - \dfrac{p^2}{2}=\dfrac{q^2}{2}+a^2$
$p^2+q^2=a^2\ (say)$
(1) becomes,
$a^2y=qz$
$q=\dfrac{a^2y}{z}$
$p=\dfrac{a}{2}\sqrt{z^2-a^2y^2}$
Since $dz=pdx+qdy$
$dz=\dfrac{a}{z}\sqrt{z^2-a^2y^2\ }dx+\dfrac{a^2y}{z}dy$
$zdz-a^2ydy=a\sqrt{z^2-a^2y^2}dx$
$\dfrac{zdz-a^2ydy}{\sqrt{z^2-a^2y^2}}=adx$
$\dfrac{d(z^2-a^2y^2)}{2\sqrt{z^2-a^2y^2}}=adx$
$\sqrt{z^2-a^2y^2}=ax+b$
$\implies z^2-a^2y^2=(ax+b)^2$
Answer is:
$\implies z^2-a^2y^2=(ax+b)^2$
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