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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 $y=\pm \sqrt{x^2+c}$.
Now if we use the substitution $y=ux⟹y^{\prime }=u^{\prime }x+u,$, 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
$\begin{array}{}\text{(1)}& -\frac{1}{2}\ln |u^2-1|=\ln \left|x\right|+c\end{array}$
For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that
$\begin{array}{}\text{(2)}& \ln \left|0\right|=\ln \left|x\right|+c\end{array}$
What does equation (2) mean? $\ln \left|0\right|$ 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 $u\ne \pm 1$. Equation (1) does not hold for $u=\pm 1$
Edit 2:
Solving equation (1) for u with $u\ne \pm 1$, we arrive at the same family of equations but with $c\ne 0$. The fact that c can be zero comes from setting $u=\pm 1$