Let $s\in (0,\infty)$, $f\ $ and $g\ $ are positive measurable functions. For which $f\ $ and $g\ $ does the following equality hold: $$\int_0^sf(x)\ g(s-x)\ dx=c_1sf(s)+c_2s\ g(s)$$ Where $c_1$ and $\ c_2$ are positive constants.
When $f=\alpha>0$ and $g=\beta>0$ then: $$\alpha \beta=c_1\alpha+c_2\beta.$$ choosing $0<c_1<\beta$ and $c_2=\alpha-c_1{\alpha\over \beta}\ $ gives the required equality.
What other classes of functions satisfy the equality?
One approach is to take the Laplace Transform of both sides: as the LHS is a convolution its LT is particularly nice. First, we replace $s$ with $t$ so as to avoid confusion with the usual LT variables: \begin{align*} \mathcal{L}\Bigg\{\int_0^tf(x)\,g(t-x)\,dx&=c_1t\,f(t)+c_2t\,g(t)\Bigg\}\\ F(s)\,G(s)&=-c_1 F'(s)-c_2G'(s). \end{align*} Now this is technically a single DE with two unknowns, so it's not solvable as is. We could make an equality assumption: $F=G,$ and see what happens. We would have $$(F(s))^2=-F'(s)(c_1+c_2),$$ with solution $$F(s)=\frac{c_1+c_2}{s-(c_1+c_2) C}, $$ so that $$f(t)=(c_1+c_2)\,e^{(c_1+c_2)Ct}. $$ This is a positive, measurable function, and you can plug this into the original functional equation for both $f$ and $g$ and see that it works.
There is no guarantee that these are the only solutions.