I've been going through my book learning about differential equations of multiple variables and I have a quick question about differential forms.(adsbygoogle = window.adsbygoogle || []).push({});

If you are working a problem and get to the point where you're left with a differential form like ##(y)dx##, does that mean that the change in the function as ##x## changes is zero, such that ##(y)dx=0##?

For example, one of the example problems in my book is:$$(2x^2+y)dx+(x^2y-x)dy=0$$

It says that if you multiply the function by the integrating factor ##u(x)=x^{-2}## you get $$(2+yx^{-2})dx+(y-x^{-1})dy=0$$ and you end up losing the solution ##x≡0##.

I understand that plugging 0 into the original equation yields ##(y)dx+(0)dy=0##, but I'm not sure why 0 is a solution.

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# I Quick Differential Form Question

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