Differential equation - general and parictular solution

Hi!

I have been given an equation as following:

$\displaystyle y^{//}-y^{/}+\frac{1}{4}y = x-4$

Which I am to find the general $\displaystyle y^g_n$ and a particular solution $\displaystyle y^p_n$

I have found that this gives a general solution:

by

$\displaystyle y^g_n = r^2-r+\frac{1}{4}$

One solution = $\displaystyle \frac{1}{2}$

Which gives:

$\displaystyle y^g_n = Ce^\frac{x}{2} + Dxe^\frac{x}{2}$

I am then to find the particular solution $\displaystyle y^p_n$

How to proceed is then my question...

By setting:

$\displaystyle y^p_n = an+b$

and calculate from there?

How would that look?