Calculus (3rd Edition)

$$y=Ce^{-\frac{x^2}{2}}+1$$
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: $$y=Ce^{-\frac{x^2}{2}}+1$$