solve the differential equation $ y' +xy=c$ I know how to solve differential equations but I think I am getting confused because there is a constant. I began by using integrating factors but when then  end up with a very complicated integral which doesn't seem right because the original equation seems pretty straight forward. Please help!
I used the integrating factor $\mu=e^{x^2/2}.$ After multiplying the entire equation by $\mu$ and integrating I end up with $e^{x^2/2}y= \int Ce^{x^2/2}\,dx.$ 
I apologize if the formatting is off. I haven't quite gotten the hang of this syntax. 
 A: Here are the steps 
$$ \frac{d}{dx}[y]+xy=c $$
Let $\mu=\exp\left(\int x\ dx\right)= \exp\left(\frac{x^2}{2}\right)$. So now
$$ 
\mu\frac{d}{dx}[y]+\mu xy=c\mu
$$
$$ 
\exp\left(\frac{x^2}{2}\right)\frac{d}{dx}[y]+ \exp\left(\frac{x^2}{2}\right)xy=c\exp\left(\frac{x^2}{2}\right)
$$
$$ 
\exp\left(\frac{x^2}{2}\right)\frac{d}{dx}[y]+ \frac{d}{dx}\left[\exp\left(\frac{x^2}{2}\right)\right]y=c\exp\left(\frac{x^2}{2}\right)
$$
$$ 
\frac{d}{dx}\left[\exp\left(\frac{x^2}{2}\right)y\right]=c\exp\left(\frac{x^2}{2}\right)
$$
$$ 
d\left[\exp\left(\frac{x^2}{2}\right)y\right]=c\exp\left(\frac{x^2}{2}\right)\ dx
$$
$$ 
\int d\left[\exp\left(\frac{x^2}{2}\right)y\right]=c\int \exp\left(\frac{x^2}{2}\right)\ dx
$$
$$ 
\exp\left(\frac{x^2}{2}\right)y+C=c\int \exp\left(\frac{x^2}{2}\right)\ dx
$$
$$ 
\exp\left(\frac{x^2}{2}\right)y=c\int \exp\left(\frac{x^2}{2}\right)\ dx+C
$$
$$ 
y=\frac{c\int \exp\left(\frac{x^2}{2}\right)\ dx+C}{\exp\left(\frac{x^2}{2}\right)}
$$
Note that the above integral cannot be expressed in terms of elementary functions.
