Question: Suppose $$y^{'}+p(x)y=q(x),\,\,\,y(x\_0)=y\_0$$ where $p$ and $q$ are continuous functions in some interval $I$ containing $x\_0$. Show that the particular solution is $$y(x)=e^{\int\_{x\_o}^{x}{p(t)}dt}[\int\_{x\_0}^{x}{e^{\int\_{x\_0}^{t}{p(\xi)}d\xi}q(t)dt+y\_0}]$$ 
I have no idea where the $\xi$ comes from. I can only get the general solution $$y(x)=\frac{1}{I(x)}{\int{}I(x)q(x)dx+C} ,$$ where $I(x)$ is an integrating factor

Answer: 
Searching for the method called **Variation of Parameters**, we will find out, for the linear 1-order differential equation $y'+p(x)y=q(x)$ where the functions $p(x), q(x)$ have the conditions as you gave them above; there is a solution like $y\_1(x)=\text{e}^{\int-p(x)dx}$.(You know all of these)

The method goes further and tells us that the one-parameter family of solutions of our equation is as the form you noted above as well. In fact we set $y(x)=v(x)y\_1(x)$ into the equation to find another part of solution which is free of any constant. This is the particular solution $$y\_p(x)=\text{e}^{\int-p(x)dx}\int\text{e}^{\int-p(x)dx}f(x)dx$$. In 2-order linear equation we can easily understand why this approach was made.
Now I make an example to see why that formula arisen. Let we have $y'=f(x,y),\; y(x\_0)=y\_0$. You surely accept that if $f(x,y)$ be continuous in a region containing the point $(x\_0,y\_0)$ then by integrating from both sides of our latter OE, we have $$y(x)=c+\int\_{x\_0}^{x}f(t,y(t))dt$$ and certainly $$y(x\_0)=c+\int\_{x\_0}^{x\_0}f(t,y(t))dt=c$$ and so $$y(x)=y\_0 +\int\_{x\_0}^{x}f(t,y(t))dt$$. I hope you got the point. Moreover @experimentX gave you additional points.
if you multiply $e^{\int p(x)\text{d}x}$ to both sides of the equation,get:
$$y'e^{\int p(x)\text{d}x}+p(x)ye^{\int p(x)\text{d}x}=q(x)e^{\int p(x)\text{d}x}$$
the above equation can be written as
$$\text{d}\left(ye^{\int p(x)\text{d}x}\right)=\text{d}\left(\int q(x)e^{\int p(x)\text{d}x}\right)$$
then you can get the answer.