Question

p²+q²=4

Answer 1

$\frac {dx}{-fp} =\frac {dy}{-fq} =\frac {dz}{(-p*fp-q*fq)} =\frac {dp}{(fx+x*fz)} =\frac {dq}{(fy+y*fz)}\\ Since,\space\frac {dq}{(fy+y*fz)}=\frac {dq}{0}\\ =>q=a\\ \text{from the given pde, we get}\\ =>p=\pm \sqrt{4-a^2}\\ Then,\\ dz=pdx+qdy\\ dz=\pm\sqrt{4-a^2}dx+ady\\ \text{Integrating both side, we get}\\ z=\pm\sqrt{4-a^2}x+ay+c\\ \text{NOTE:- We can take either p is constant from} \frac {dp}{(fx+x*fz)}=\frac {dp}{0} \text{or qas a constant.}$