$$\begin{align} \frac{dy}{dx} &= \frac{y}{2x} \\ \frac{dy}{y} &= \frac{1}{2x} dx \\ \int\frac{1}{y} dy &=\int\frac{1}{2x} dx \\ \ln|y| &= \frac{1}{2} \ln|x| + C \\ y&=Ce^{\frac{1}{2}\ln|x|}=Ce^{\ln|x^\frac{1}{2}|}=C\sqrt{x} \end{align}$$
My book gets a different answer. Namely, $y^2=Cx$. Is there some reason why we would want to avoid a square root in our answer?
I guess it is only for aesthetical reasons because $y^2=Cx$ looks slightly prettier that $y=C\sqrt{x}$. Since your $C$ is an arbitrary one can set the square root of the one $C$ within the text book solution equal to your $C$.