Find the only periodic solution of an ODE

Question

Find the only periodic solution for $y'+y=b(x)$ with $b:\Bbb{R}\to\Bbb{R}$ has a period of $2T$ and is $1$ for $x (0,T)$ and $-1$ for $x (-T,0)$.

The ODE is easy to solve: $y(x) = \exp(-x)\cdot c+1$ and $y(x) = \exp(-x)\cdot c-1$. But how can I find the $c$ such that the solution is periodic with a period of $2T$?

The solution is:

Answer 1

Any ODE solution that deserves its name is continuous. So you need to establish continuity at $x=0$ and $y(-T)=y(T)$ to have a periodic solution. For the first you need $$ 1+c_+=-1+c_-\implies c_-=2+c_+ $$ and for the second condition $$ 1+c_+e^{-T}=-1+c_-e^T=-1+(2+c_+)e^T\implies c_+\sinh(T)=1-e^T $$ etc.

Answer 2

We have

$$ \cases{ y'(t)+y(t)=1\\ -y'(\eta)+y(\eta)=-1, \ \ \ \text{with}\ \ \ \eta = -t } $$

and solving we have

$$ y_+(t) = 1+C_+ e^{-t}\\ y_-(\eta) = -1+C_- e^{\eta} $$

at $t = 0$ for continuity we have

$$ y_+(0)=y_-(0) \Rightarrow 1+C_+ = -1 + C_- $$

and for periodicity and continuity

$$ y_+(T) = -y_-(-T)\Rightarrow 1+C_+ e^{-T} = -1+C_- e^T $$

so the constants $C_+,C_-$ are determined from

$$ \cases{ 1+C_+ = -1 + C_-\\ 1+C_+ e^{-T} = -1+C_- e^T } $$