#### Answer

$$ y =\pm\sqrt{x^2+2c}.$$

#### Work Step by Step

We have
$$ yy'= x\Longrightarrow ydy=xdx\\
\Longrightarrow\int ydy=\int xdx+c\\
\Longrightarrow \frac{1}{2}y^2=\frac{1}{2}x^2+c
\Longrightarrow y =\pm\sqrt{x^2+2c}.$$
Hence, the general solution is
$$ y =\pm\sqrt{x^2+2c}.$$