Question

The primitive of the partial differential equation
√p + √q = 2x , is 1/6×(2x-a)^3+a^2y+b
True or false with full explanation

Answer 1

$\displaystyle\textsf{True}\\ \sqrt{p} + \sqrt{q} = 2x\\ \sqrt{q} = 2x - \sqrt{p} = a\\ \sqrt{q} = a, q = a^2\\ 2x - \sqrt{p} = a\\ \sqrt{p} = 2x - a\\ p = (2x - a)^2\\ \mathrm{d}z = p\,\mathrm{d}x + q\,\mathrm{d}y\\ \mathrm{d}z = (2x - a)^2\,\mathrm{d}x + a^2\,\mathrm{d}y\\ \int \,\mathrm{d}z = \int (2x - a)^2\,\mathrm{d}x + \int a^2\,\mathrm{d}y\\ z = \frac{(2x - a)^3}{6} + a^2y + b\\ \textsf{Therefore, it is true.}$