Find the only periodic solution for π¦β²+π¦=π(π₯) with π:βββ has a period of 2π and is 1 for π₯(0,π) and β1 for π₯(βπ,0).
The ODE is easy to solve: π¦(π₯)=exp(βπ₯)β π+1 and π¦(π₯)=exp(βπ₯)β πβ1. But how can I find the π such that the solution is periodic with a period of 2π?
The solution is...