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We have ${y}^{\prime }=x/y$, which is a first-order homogeneous differential equation.
It can be solved by rearranging to y dy=x dx and then integrating both parts which yields that $y=±\sqrt{{x}^{2}+c}$.
Now if we use the substitution $y=ux\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }={u}^{\prime }x+u,$, and rewrite the differential equation as
${u}^{\prime }x+u=\frac{1}{u}$
and then rearrange to
$\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$
by integrating both parts we get that
$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}|x|+c\end{array}$
For $y=±x$ (a special solution for c=0) $\to u=±1$, and by plugging $±1$ into (1) we get that
$\begin{array}{}\text{(2)}& \mathrm{ln}|0|=\mathrm{ln}|x|+c\end{array}$
What does equation (2) mean? $\mathrm{ln}|0|$ is undefined. Is this of any significance?
Edit 1:
As pointed out when rearranging from ${u}^{\prime }x+u=\frac{1}{u}$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac{1}{x}\right)dx$, we implicitly assumed that $u\ne ±1$. Equation (1) does not hold for $u=±1$
Edit 2:
Solving equation (1) for u with $u\ne ±1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=±1$
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Cristal Obrien
When you rearranged your equation to get
$\left(\frac{1}{1/u-u}\right)du=\left(1/x\right)dx$
you have implicitly assumed that $u\ne ±1$ so $u=±1$ is out of question.