$y'' - y = \frac{1}{2}e^x$
To find the general solution of the above, I understand that we need both $y_{h}(x)$ and $y_{p}(x)$.
I have found $y_{h}(x)$, the homogeneous general solution to be
$y_{h}(x) = c_{1}e^x + c_{2}e^{-x}$
And in finding $y_{p}(x)$ I have subbed $y = ue^x$ where $u = u(x)$ and derived at $u'' + 2u' = \frac{1}{2}$
However, I do not know how to proceed from here, guidance will be much appreciated, thank you!
The method you are using to find $y_{p}(x)$ is called the 'method of undetermined coefficients'.
There are lots of resources describing this method, online and in text books, and the choices for the particular integral.
See for example this, which is just one of the first I found, that describes the complication you have encountered. ( There are of course many other ways of determining a particular integral.)