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 $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, 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

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ 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 $u=\pm 1$

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 $y=ux\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}={u}^{\prime}x+u,$, 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

$\begin{array}{}\text{(1)}& -\frac{1}{2}\mathrm{ln}|{u}^{2}-1|=\mathrm{ln}\left|x\right|+c\end{array}$

For $y=\pm x$ (a special solution for c=0) $\to u=\pm 1$, and by plugging $\pm 1$ into (1) we get that

$\begin{array}{}\text{(2)}& \mathrm{ln}\left|0\right|=\mathrm{ln}\left|x\right|+c\end{array}$

What does equation (2) mean? $\mathrm{ln}\left|0\right|$ 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 $u=\pm 1$