Can I do this when integrating 1/x to get rid of the absolute value?

Question

I'm studying PDEs from a textbook and what comes up all the time is an ODE containing the integration of $1/x$. Now, if we are not sure that x is positive we can't get away from the absolute value in the integration. So, up to this point I would do this :

$$\frac{dx}{x}=dy$$ $$\ln|x|=y+C_1$$

where $C_1$ is a constant. The author,however, in such cases, does the following:

$$\frac{dx}{x}=dy$$ $$=>\ln (C_1x)=y$$

I know how you can get the constant inside the log but how does the absolute value vanish? In this problem nothing is stated about the value of x and the log includes no absolute value.

Answer 1

Since the $C_1$ can be positive or negative, it can absorb the sign on $x$. One just has to make sure that $C_1$ and $x$ have the same sign.

I would have taken a different approach. From

$$\ln |x|= y+C$$

exponentiate both sides to get

$$| x | = e^{y+C} = =e^ye^C = Ae^y$$

where $A=e^C$ is a positive constant. Then

$$x = \pm Ae^y$$

so all we need to do is allow $A$ to be positive or negative (or zero) and it "absorbs" the plus-or-minus sign. Now you have

$$x= Ae^y$$

with no absolute values. Now you can put it in the author's form, if you like.

Answer 2

$$\ln|x|=y+C_1\implies\ln|x|-C_1=y$$Here if you write $C_1'=e^{C_1}$, you have $\ln|x|-\ln (C_1')=y.$ By the quotient law of logarithm, $$\ln\dfrac{|x|}{C_1'}=y$$Note that $C_1'=e^{C_1}>0$. You can then define $D=\dfrac{1}{C_1'}.$ The original equation becomes $$\ln D|x|=y,$$for some non-zero $D$. We don't know your question, but one can do $$\begin{cases}D&=-D&\text{if}~x<0\\D&=D&\text{if}~x>0\end{cases}$$such that the absoulte sign can be taken away.