Answer to Question #137100 in Differential Equations for jahnavi

Question

P^2z+q^2=1

Answer 1

This is of the form $f(z, p, q)=0.$ Let $q=ap.$ f (z,p,q) = 0

The given  equation becomes

$$p^2z+a^2p^2=1$$

$$p=\pm\dfrac{1}{\sqrt{z+a^2}}$$

$$q=\pm\dfrac{a}{\sqrt{z+a^2}}$$

Since $dz=pdx+qdy,$

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

$$\sqrt{z+a^2}dz=\pm(dx+ady)$$

On integration

$$\dfrac{2}{3}(z+a^2)^{3/2}=x+ay+b$$

$$(z+a^2)^{3/2}=\dfrac{3}{2}x+\dfrac{3}{2}ay+b$$