I try to find the general solution of the $$y'(x)+4y'(\frac{x}{2})+1=0$$ as follow:
1-the complementary solution $$y'(x)+4y(\frac{x}{2})=0$$ $$y_c=e^{rx}$$ so $$y'(x)=re^{rx}$$ $$y'(\frac{x}{2})=\frac{r}{2}e^{\frac{rx}{2}}$$ then $$re^{rx}+2re^{\frac{rx}{2}}=0$$ $$re^{\frac{rx}{2}}(e^{\frac{rx}{2}}+2)=0$$ and what after that ???
Note that taking the derivative of $$ y(x)+8y(\frac{x}{2}) + x + C = 0 $$ lead to your equation.
Try $y(x) = ax + b$ then $$ (ax + b) + 8(a\frac{x}{2} + b) + x + C = 0 $$ which is the same as $$ (5a + 1)x + (9b + C) = 0 $$ Now this lead to $$ 5a + 1 = 0 $$ and $a = -1/5$ and likewise $b = - C/9$ now $C$ is arbritrary and hence we can write a solution as $$ y(x) = D - \frac{x}{5}, $$ with arbrirary constant $D$