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I am asking because i think we divided by x here for whatever reason since the other side is equal to 0 and it wont affect the equation in any meaningful way.enter image description here

Letting $y=ux$ we have $$\begin{align} (x-ux) dx + x(udx + x du) &= 0 \\ dx+ x du &= 0\\ \frac{dx}x+du&=0\\ \ln |x| + u &= c\\ x\ln |x|+ y &= cx. \end{align}$$

I got $x/2 = y + c$ instead

The thing I don't understand though is how this is legal, because if it is it means there is an infinite number of solutions possible as we could also multiply both sides by anything.

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  • $\begingroup$ I see that you have created. I think it is worth pointing out that this tag is now being discussed on meta. $\endgroup$ Mar 14, 2015 at 8:31
  • $\begingroup$ From FAQ about tags: Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag. It's not easy to keep balance between too specific tags and not having enough tags, but it is always good to search first and to ask yourself, whether newly created tag is not too specific. Another thing - when creating a tag, it is useful to create a tag-wiki when somebody creates a new tag. $\endgroup$ Mar 14, 2015 at 8:32

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Yes, provided that term is non-zero.

Notice that after division by $x$, and a non-explicit integration, we had a $\ln |x|$ term which, just like $\frac{1}{x}$, is not well-defined when $x=0$.

Notice that several lines have been missed out. Going from the third to the fourth line, we had:

\begin{array}{ccc} \frac{\operatorname{d}\!x}{x} + \operatorname{d}\!u &=& 0 \\ \\ \frac{\operatorname{d}\!x}{x} &=& -\operatorname{d}\!u \\ \\ \int \frac{\operatorname{d}\!x}{x} &=& -\int \operatorname{d}\!u \\ \\ \ln|x| &=& -u + c \\ \\ \ln|x| + u &=& c \end{array}

The original question was in terms of $y$ and $y$ was swapped for $ux$. If $y=ux$ then $u=\frac{y}{x}$, so lets make the swap:

$$\ln|x| + \frac{y}{x} = c$$.

Since $x \neq 0$, we can multiply through by $x$ to give:

$$x\ln|x| + y = cx$$

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  • $\begingroup$ So there is an infinite number of solutions (integrals) possible? $\endgroup$ May 14, 2013 at 19:54
  • $\begingroup$ @GladstoneAsder Yes. The constant of integration allows for an uncountably infinite set of solution. Pick a different number $c$ and you get a different solution. $\endgroup$ May 14, 2013 at 19:56
  • $\begingroup$ can we multiply both sides by y? $\endgroup$ May 14, 2013 at 19:59
  • $\begingroup$ also i just realized what i did wrong, i treated x^2 as a constant when i should have put it to the other side. basically, they did the same thing. so, can we treat x^2 as a constant and integrate x^2 for u? $\endgroup$ May 14, 2013 at 20:01
  • $\begingroup$ I've updated my original reply with a more detailed explanation $\endgroup$ May 14, 2013 at 20:05

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