Question

yp+x^2q^2=2x^2y

Answer 1

$$yp+x^2q^2=2x^2y\\ \frac{yp}{2x^2y}+\frac{x^2q^2}{2x^2y} =\frac{2x^2y}{2x^2y}\\ \frac{p}{2x^2}+\frac{q^2}{2y} =1\\ \frac{q^2}{2y} =1-\frac{p}{2x^2}=a \qquad \text{(say)}\\ \therefore \qquad \frac{q^2}{2y}=a \implies q = \sqrt{2ay}\\ \text{and } \qquad 1-\frac{p}{2x^2}=a \implies p =2x^2(1-a)$$

Since:

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

Then:

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

Integrating through:

$$\int dz = \int 2x^2(1-a)\;dx + \int \sqrt{2ay}\;dy\\ z = \frac{2x^3(1-a)}{3} + \frac{\sqrt{8ay^3}}{3}+c\\ 3z = 2x^3(1-a) + \sqrt{8ay^3} + b\\$$

where b=3c