How bad can perturbing a functional equation really make things? A long time ago, I found occassion to find solutions to a functional equation of the following form 
$f(x-y) = f(x) - f(y) + \delta $ with $\delta \in \mathbb{R}.$
Using the same exact techniques as in solving the case where $\delta =0,$ you can show that the family of solutions on the rationals is given by $f_{\delta } (x) = \chi_{\mathbb{R}_+ }(x) (f(1) - \delta ) x +\delta ,$ with the same Hamel basis argument going through as usual to show the general solution. 
It seems worthwhile to try this technique out elsewhere. For the equation $f(xy) = f(x)f(y) + \delta ,$ however, things already seem more complicated - for instance, we must have that $f(0), f(1) = 1 \pm \sqrt{ 1 - 4\delta }$ and that $f(x^m) = \displaystyle\sum_{k=2}^{\lfloor m/2 \rfloor} \delta  f(x)^{m-2k} .$ It seems that the usual tricks aren't going to work. My question, as it pertains to this and other examples, is whether or not there is something akin to reducing functional equations to a homogenous case, and as whether or not there are other methods such problems are amenable to.
 A: For the equation $f(x-y)=f(x)-f(y)+ \delta$, simply let $g(x)=f(x)- \delta$, then the given equation becomes $g(x-y)=g(x)-g(y)$. (The same equation but with $\delta =0$)
For the equation $f(xy)=f(x)f(y)+ \delta$, where $\delta \not =0$:
Substituting $y=0$ gives $f(0)=f(x)f(0)+ \delta$.
Since $\delta \not =0, f(0) \not =0$, so $f(x)=\frac{f(0)-\delta}{f(0)}=c$, for some constant $c$. Putting this into the original equation we get $c=c^2+ \delta$, so that $c=\frac{1 \pm \sqrt{1-4 \delta}}{2}$. (Note: if you do not restrict the domain of $f(x)$ to be $\mathbb{R}$, all $\delta$ work, while if you did restrict to $\mathbb{R}$, you would require $\delta \leq \frac{1}{4}$.)
If you add a parameter $\delta$ to the equation, typically 1 of 2 things happen:


*

*You can find a suitable transformation of $f(x)$ to eliminate $\delta$, e.g. $g(x)=f(x)- \delta$ in the equation $f(x-y)=f(x)-f(y)+ \delta$. For the equation $f(xy)=f(x)f(y) \delta$ suggested in the comment, just use $g(x)=f(x) \delta$ for $\delta \not =0$. (For $\delta =0$, clearly $f(x)=0 \, \forall x$)

*The introduction of $\delta \not =0$ allows you to exploit $\delta$ to get further restrictions for the function, as I did for the equation above. Note that $f(xy)=f(x)f(y)$ has continuous solutions $x^a$ (and many more which are discontinuous, similar to the Cauchy equation), while $f(xy)=f(x)f(y)+ \delta$ only has solutions $f(x)=\frac{1 \pm \sqrt{1-4 \delta}}{2}$.

