Question

p-3q = sinx+cosy

Answer 1

$$p-3q = \sin x+\cos y$$

$$p- \sin x=3q +\cos y=a (say),$$

where $a$ is an arbitrary constant.

We have

$$\begin{cases} p-\sin x=a \\ 3q+\cos y=a \end{cases}=> =\begin{cases} \dfrac{\partial z}{\partial x}=a+\sin x \\ \dfrac{\partial z}{\partial y}=\dfrac{1}{3}(a-\cos y) \end{cases}$$

$$dz=(a+\sin x)dx+(\dfrac{1}{3}(a-\cos y))dy$$

$$z=ax-\cos x+\dfrac{1}{3}ay-\dfrac{1}{3}\sin y-b$$

Or

$$z=a(x-\dfrac{1}{3}y)-(\cos x+\dfrac{1}{3}\sin y+b)$$