Homogeneous first-order Differential Equation $y' = x/y$

Question

We have $y' = x/y$, 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 $y=ux\implies y'=u'x+u,$ and rewrite the differential equation as $$u'x+u=\frac1u$$ and then rearrange to $$\left(\frac{1}{1/u-u}\right)du=\left(\frac1x\right)dx$$ by integrating both parts we get that $$-\frac{1}{2}\ln\left |u^2 -1 \right| = \ln\left |x \right |+c \tag1$$

For $y=\pm x$ (a special solution for $c=0$) $\rightarrow u=\pm1$, and by plugging $\pm1$ into $(1)$ we get that $$\ln\left |0 \right|=\ln\left |x \right| +c \tag2$$

What does equation $(2)$ mean? $\ln\left |0 \right|$ is undefined. Is this of any significance?

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Edit 1

As pointed out when rearranging from $u'x+u=\frac1u$ to $\left(\frac{1}{1/u-u}\right)du=\left(\frac1x\right)dx$, we implicitly assumed that $u\neq\pm1$. Equation $(1)$ does not hold for $u=\pm1$.

Edit 2

Solving equation $(1)$ for u with $u\neq\pm1$, we arrive at the same family of equations but with $c\neq0$. The fact that $c$ can be zero comes from setting $u=\pm1$.

Answer 1

When you rearranged your equation to get $$(\frac{1}{1/u-u})du=(1/x)dx$$ you have implicitly assumed that $u \ne \pm 1$ so $u=\pm 1$ is out of question.

Answer 2

Note that for the substitution $y=ux$ $$u'x+u=\frac 1 u \implies \int \frac {u}{1-u^2}du=\int \frac {dx}x$$

Then for $( u\neq \pm1,0)$

$$-\frac 12 \int \frac {-2u}{1-u^2}du=\int \frac {dx}x \implies 1-u^2=\frac K{x^2}$$ $$\ln(1-u^2)=-2\ln(x)+K \implies 1-u^2 =e^{-2\ln(x)+K}$$ $$\implies 1-u^2=\frac K{x^2}$$ $$\implies x^2-y^2=K$$ For the value $1,-1$ ( 0 is excluded) you can use the original equation

$$u'x+u=\frac 1u$$

Edit

My post corrected some mistakes Op made in the original question...Since then the question has changed.