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