How do we deal with modulus signs when finding the solution of a differential equation? Consider: $\quad y'=\frac{y}{2x} \quad $ where we're required to find the general solution in the form $y=y(x)$
$\quad y'=\frac{y}{2x} \quad \rightarrow \quad \int \frac{1}{y} dy=\int \frac{1}{2x} dx$
$\hspace{2.5cm} \rightarrow \quad \ln|y|=\frac{1}{2}\ln|x|+C $
$\hspace{2.5cm} \rightarrow \quad \ln|y|=\ln\sqrt{|x|}+C  \quad[1]\quad or \quad \ln{y^2}=\ln|x|+C \quad [2]$
For [1] I can continue as  follows:
$\ln|y|=\ln{C\sqrt{|x|}} \quad \rightarrow \quad y=C\sqrt{|x|} \quad$ 
as the constant C accounts for the $\pm$ that would normally be required $y=\pm C\sqrt{|x|}$
From [2] we have:
$\ln{y^2}=\ln{C|x|} \quad \rightarrow y^2=C|x| \quad \rightarrow \quad y=\pm \sqrt{C|x|}=\pm C \sqrt {|x|}=C\sqrt{|x|} $
To the best of my knowledge, I believe that this is the simplest form I can give my solution in; I wanted to know if its at all possible to remove the modulus sign and still have a solution in the form $y=y(x)$?

You can always write
$$
\frac{y(x)}{y_0}=\sqrt{\frac{x}{x_0}},
$$
as in any solution, there can be no sign change in neither $x$ nor $y$. Thus the fractions are always positive. Along the way this also takes care of the integration constant by directly expressing it in the initial conditions.
