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.

  • $\begingroup$ 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
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1 Answer 1


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:

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

  • $\begingroup$ 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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