We have ODE $$0=-\frac{x}{y}+\frac{y}{x}y'$$ where $(x,y)\in]0,\infty[\times]0,\infty[$.
It's inexact, since$$\frac{dP(x,y)}{dy}=\frac{x}{y^2}\neq-\frac{y}{x^2}=\frac{dM(x,y)}{dx}$$
I can only find one integrating factor $R(x,y)=xy$, then we get $$\frac{dP(x,y)\cdot{R(x,y)}}{dy}=-\frac{dx^2}{dy}=0=\frac{dy^2}{dx}=\frac{dM(x,y)\cdot{R(x,y)}}{dx}$$
Then I can solve this problem correctly. What I am not quite sure about is are we allowed to get $0$ as result after derivation at x and y. And is there another way to find a better integrating factor?
Thanks in advance!
$$-\frac{x}{y}+\frac{y}{x}y'=0$$ The integrating factor is $\mu(xy)=xy$: $$-{x^2}dx+{y^2}dy=0$$ Then the DE is exact since: $$\partial_y (-x^2)=\partial_x(y^2)=0$$ Then integrate: $$-{x^2}dx+{y^2}dy=0$$ $${x^3}-{y^3}=C$$
$$P(x,y)dx+Q(x,y)dy=0$$ When you have: $$\partial_y (P(x,y))=0 \implies P=P(x)$$ $$\partial_x (Q(x,y))=0 \implies Q=Q(y)$$ the DE is exact anf also separable: $$P(x)dx+Q(y)dy=0$$ $$\int P(x)dx=-\int Q(y)dy$$