Is there a continuous function g that satisfies $2sg'(s)=g(1-s) - g(s)$ Is there a continuous function g that satisfies $$2sg'(s)=g(1-s)  -  g(s)$$
Could you show how to arrive at the solution 

Hint: 


*

*try to find a higher degree ODE by derivation of the equation.

*then notice that you can substitute $s$ by $1-s$ in the equation to express $g'(1-s)$ in function of $g'(s)$.

*you are left with $\dfrac{g''(s)}{g'(s)}=f(s)$
Solution $$g(s)=a\sqrt{\pm\frac{s-1}{s}}+b$$
You cannot have continuity in $0$ (well, except for constant functions $a=0$) however it is possible in $1$ by selecting $b_{[0,1]}=b_{[1,+\infty)}$ and relative smoothness by selecting $a_{[0,1]}=-a_{[1,+\infty)}$ (still a vertical tangent though...).
