Suppose we have a differential equation like this: $$P(x,y)\operatorname{d} \!x + Q(x,y)\operatorname{d}\!y = 0 \quad \textrm{(1)}$$ We say that if the equation is exact, i.e, there is a $\psi(x,y)$ such that $\frac{\partial \psi}{\partial x} = P$ and $\frac{\partial \psi}{\partial y} = Q$ then $\psi(x,y) = c$ is a solution. Basically what we're doing here is finding a level surface, whose incliniation (the vector $\{dx,dy\}$) will always be orthogonal to the vector field defined by $\vec{F} = P(x,y)\vec{i} + Q(x,y)\vec{j}$.

Then we have the integrating factor. Suppose that equation $\textrm{(1)}$ is not exact, and we use a integrating factor to find a function $\psi$. Why the integrating factor won't change the vector field? If it changes, how can this solution be the same than the one for the original vector field?

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    $\begingroup$ The integrating factor changes the vector field, but does not change the solution since it does not change the differential equation. $\endgroup$ – shrinklemma Apr 17 '17 at 3:08
  • $\begingroup$ If it changes the vector field, then the solution is a curve that is orthogonal to both the original and the second vector field? $\endgroup$ – Vitor C Goergen Apr 17 '17 at 3:09
  • $\begingroup$ Yes, since the two vector fields are in the same direction at each point, orthogonality of the curve to one of the vector fields is equivalent to orthogonality to the other one. $\endgroup$ – shrinklemma Apr 17 '17 at 3:56
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    $\begingroup$ How can i verify that the integrating factor doesn't change the directions of the vector field? $\endgroup$ – Vitor C Goergen Apr 17 '17 at 12:31

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