Question

(1-x^2)yp^2+x^2q=0

Answer 1

$$(1-x^2)yp^2+x^2q=0\\ \frac{(1-x^2)yp^2}{p^2q}+\frac{x^2q}{p^2q}=\frac{0}{p^2q}\\ \frac{(1-x^2)y}{q}+\frac{x^2}{p^2}=0\\ -\frac{(x^2-1)y}{q}+\frac{x^2}{p^2}=0\\ \frac{x^2}{p^2}=\frac{(x^2-1)y}{q}=a \qquad \text{(say)}\\$$

Then:

$$\frac{(x^2-1)y}{q}=a \implies q= \frac{(x^2-1)y}{a}\\ \frac{x^2}{p^2}=a \implies p = \frac{x}{\sqrt{a}}$$

Since

$$dz = \dfrac{\partial z}{\partial x}dx + \dfrac{\partial z}{\partial y}dy$$

Therefore,

$$dz = \frac{x}{\sqrt{a}}dx+\frac{(x^2-1)y}{a}dy$$

Integrating through:

$$z = \frac{x^2}{2\sqrt{a}}+\frac{(x^2-1)y^2}{2a}+c\\ \therefore 2z = \frac{x^2}{\sqrt{a}}+\frac{(x^2-1)y^2}{a}+k\\$$

where $k = 2c$