Quick Differential Form Question

[Drakkith]

I've been going through my book learning about differential equations of multiple variables and I have a quick question about differential forms.

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.

 

[fresh_42]

$ydx$ isn't automatically zero, but the system ( $x\equiv 0 \wedge ydx=0$ ) has a solution which will get lost if we divide by $x^{2}$. So as always in these cases, consider this system first and in the next case $x \not\equiv 0\,$.

 

[Drakkith]

$ydx$ isn't automatically zero, but the system ( $x\equiv 0 \wedge ydx=0$ ) has a solution which will get lost if we divide by $x^{2}$.

I'm sorry but I don't know what that bit in the parentheses means.

So as always in these cases, consider this system first and in the next case $x \not\equiv 0\,$.

What do you mean by "consider this system first"?

 

[fresh_42]

I meant $x \equiv 0$ is a possibility at first. In this case we conclude $ydx=0$ which can be integrated and leads to a solution with $x \equiv 0$ and $y = \text{ anything }$. From this point on we can assume $x \not\equiv 0$ and consider this remaining case.

 

[Drakkith]

Ah okay. Thanks fresh.

 

[WWGD]

As a general comment not every operation on your equation/system will preserve the solution set. Remember, e.g., when you have a system of linear equations that anything other than switching rows, non-zero scaling and adding a multiple of one row to another, will change the solution .