How to solve this pde equation: $(p^2 + q^2)y = qz$

Question

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: $z^2=(cx+a)^2 + c^2 y^2$ .

Answer 1

Hint:

Let $z=e^u$ ,

Then $z_x=e^uu_x$

$z_y=e^uu_y$

$\therefore((e^uu_x)^2+(e^uu_y)^2)y=e^uu_ye^u$

$ye^{2u}u_x^2+ye^{2u}u_y^2=e^{2u}u_y$

$yu_x^2+yu_y^2=u_y$

Answer 2

$let\ f(x,y,z,p,q)=(p^2+q^2)y-qz=0 \hspace{1cm} \to (1)$

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:

$\implies z^2-a^2y^2=(ax+b)^2$