Question

z^2(p^2+q^2+1)=1 solve this partial differential equation​

Answer 1

given,

$z^2(p^2+q^2+1)=1 \\differential\\ equation\\\mathrm{Divide\:both\:sides\:by\:}z^2;\quad \:z\ne \:0\\\frac{z^2\left(p^2+q^2+1\right)}{z^2}=\frac{1}{z^2};\quad \:z\ne \:0\\p^2+q^2+1=\frac{1}{z^2}\\\mathrm{Subtract\:}q^2\mathrm{\:from\:both\:sides}\\p^2+q^2+1-q^2=\frac{1}{z^2}-q^2\\p^2+1=\frac{1}{z^2}-q^2\\p^2=\frac{1}{z^2}-q^2-1\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\p=\sqrt{\frac{1}{z^2}-q^2-1}$

$p=\frac{\sqrt{1-z^2q^2-z^2}}{z}$