Answer to Question #219107 in Differential Equations for Tohid

Question

 z2(p2 + q2 + 1) = k2

Answer 1

$$f(x, y, z, p, q)=z^2(p^2+q^2+1)-k^2=0$$

$$f_p=2pz^2, f_q=2qz^2,$$

$$f_x=0, f_y=0, f_z=2z(p^2+q^2+1)$$

Charpit's Method

$$\dfrac{dx}{f_p}=\dfrac{dy}{f_q}=\dfrac{dz}{pf_p+qf_q}=\dfrac{dp}{-(f_x+pf_z)}=\dfrac{dq}{-(f_y+qf_z)}$$

Then

$$\dfrac{dp}{-(0+2pz(p^2+q^2+1))}=\dfrac{dq}{-(0+2qz(p^2+q^2+1))}$$

$$\dfrac{dp}{p}=\dfrac{dq}{q}$$

$$\int\dfrac{dp}{p}=\int\dfrac{dq}{q}$$

$$\ln|p|=\ln|q|+\ln a$$

$$p=qa$$

Substitute

$$z^2(a^2q^2+q^2+1)=k^2$$

$$q^2=\dfrac{k^2-z^2}{z^2(a^2+1)}$$

$$q=\sqrt{\dfrac{k^2-z^2}{z^2(a^2+1)}}$$

$$p=a\sqrt{\dfrac{k^2-z^2}{z^2(a^2+1)}}$$

$$dz=a\sqrt{\dfrac{k^2-z^2}{z^2(a^2+1)}}dx+\sqrt{\dfrac{k^2-z^2}{z^2(a^2+1)}}dy$$

$$z\sqrt{\dfrac{a^2+1}{k^2-z^2}}dz=adx+dy$$

Integrate

$$-\sqrt{a^2+1}\sqrt{k^2-z^2}=ax+y+b$$

$$\sqrt{a^2+1}\sqrt{k^2-z^2}+ax+y+b=0$$