Answer
$$y= x^{-1}\left( \frac{1}{2} \sin(x^2)+C\right) .$$
Work Step by Step
This is a linear equation which has the integrating factor as follows
$$\alpha(x)= e^{\int P(x)dx}=e^{\int x^{-1} dx}=e^{\ln x}=x$$
Now the general solution is
\begin{align}
y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x^{-1}\left( \int x\cos(x^2)dx +C\right)\\ &=x^{-1}\left( \frac{1}{2} \sin(x^2)+C\right) \end{align}
The general solution is
$$y= x^{-1}\left( \frac{1}{2} \sin(x^2)+C\right) .$$
