Sorry for the provocative question but I am often see a solution where the absolute value is neglect for example:
$$ \begin{cases} xuu_x+yuu_y=u^2-1, x>0\\ u(x,x^2)=x^3\\ \end{cases} $$
We look at:
$$\frac{dx}{xu}=\frac{dy}{yu}=\frac{du}{u^2-1}$$
We solve: $$ \begin{cases} \frac{dx}{xu}=\frac{dy}{yu}\\ \frac{dx}{xu}=\frac{du}{u^2-1}\\ \end{cases}$$
for $$\frac{dx}{xu}=\frac{dy}{yu}$$ we can multiple by $u\neq 0$:
$$\frac{dx}{x}=\frac{dy}{y}$$
which is:
$$\ln(x)=\ln|y|+c^*_1\rightarrow \ln|y|-\ln(x)=c^*_1\rightarrow \ln(\frac{|y|}{x})=c^*_1\rightarrow \frac{|y|}{x}=c_1$$
but on the recitation answer the solved:
$$\frac{dx}{x}=\frac{dy}{y}\rightarrow \ln(x)=\ln(y)+c^*_1$$
Why can this be done? What am I missing?
Technically speaking, although the expression $\displaystyle\int\frac1x\,dx=\ln|x|+C$ is correct as far as it goes, there is a discontinuity at $x=0,$ requiring the flexibility of different constants of integration. This integral is really only defined on an interval not containing zero. So what you should really write is this: $$\int\frac1x\,dx=\begin{cases}\ln(x)+C_1, & x>0\\ \ln(-x)+C_2,\quad & x<0 \end{cases}. $$ We can check that this is correct by differentiating each piece. For $x>0,$ we have $$\frac{d}{dx}[\ln(x)+C_1]=\frac1x. $$ For $x<0,$ we have $$\frac{d}{dx}[\ln(-x)+C_2]=\frac{1}{-x}\,(-1)=\frac1x, $$ as required.