Answer
$y=\sqrt{2(\ln(x)+2)}$
Work Step by Step
Using the FTC it follows:
$$y'=\frac{1}{xy}$$
$$ydy=\frac{1}{x}dx$$
$$\int ydy=\int \frac{1}{x}dx$$
$$\frac{y^{2}}{2}=\ln(x)+C$$
$$y^{2}=2(\ln(x)+C)$$
From the given equation for $x=1$, $y=2+0=2$
$$y^{2}=2(\ln(x)+C)$$
$$2^{2}=2(\ln(1)+C) \to C=2$$
$$y^{2}=2(\ln(x)+2)$$
$$y=\sqrt{2(\ln(x)+2)}~~\text{or}~~y=-\sqrt{2(\ln(x)+2)}$$
The equation $y=-\sqrt{2(\ln(x)+2)}$ should be discarded because when $x=1$, the value of $y$ is $-2$ which is not adequat with the initial condition. So the solution is:
$$y=\sqrt{2(\ln(x)+2)}$$
