Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) https://en.m.wikipedia.org/wiki/Functional_differential_equation
– JepsilonJan 11 '19 at 07:50
1
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=\frac{Ae}{2}$. Then you find $A$ with some initial condition.
– JepsilonJan 11 '19 at 08:06