If the solution to the differential equation $P(x,y)dx + Q(x,y)dy =0$ is $\Phi(x,y)=c,$ then that would mean that $$\frac{\partial\Phi}{\partial y}= Q(x,y)\hspace{1cm} (1)$$ and $$\frac{\partial\Phi}{\partial y}= P(x,y)\hspace{1cm} (2)$$ My ODE textbook loses me in the next paragraph where it states that it is difficult for equations $(1)$ and $(2)$ to be true for any random $Q$ and $P$ and that is why we use the $$\partial P/\partial y=\partial Q /\partial x \hspace{1cm}(4)$$which is not true for any $P$ and $Q$ and if $(4)$ is true then the equation is exact.
Could someone please explain this part? Thank you in advance.
To integrate a differential combination of the form $$P\mathrm dx+Q\mathrm dy,$$ we need to find a family of functions $\phi$ so that the differential $\mathrm d\phi$ of $\phi$ is equal to the linear combination above.
But to differentiate a function $f$ of two variables $x,y,$ we know from the differential calculus that $$\mathrm df=f_x\mathrm dx+f_y\mathrm d y.$$ Thus if the form $P\mathrm dx+Q\mathrm dy$ is to be equal to some differential $\mathrm d\phi,$ it must be the case that $P=\phi_x$ and $Q=\phi_y.$ Now we know that when the first two derivatives of a function are continuous, then the mixed derivatives are equal. Thus, it is the case that $$P_y=\phi_{yx} =\phi_{xy}=Q_x,$$ so that then we will be able to arrange the form in easily factored form, or integrate in two different ways, to determine $\phi$ up to a constant quantity.