Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 23

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

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