The ODE is separable.
[tex]\dfrac{dy}{dx} = xy \iff \dfrac{dy}y = x\,dx[/tex]
Integrate both sides to get
[tex]\displaystyle \int\frac{dy}y = \int x\,dx[/tex]
[tex]\boxed{\ln|y| = \dfrac12 x^2 + C}[/tex]
But notice that replacing the constant [tex]C[/tex] with [tex]-C[/tex] doesn't affect the solution, since its derivative would recover the same ODE as before.
[tex]\ln|y| = \dfrac12 x^2 - C \implies \dfrac1y \dfrac{dy}{dx} = x \implies \dfrac{dy}{dx} = xy[/tex]
so either of the first two answers are technically correct.
