Differential equation of period Suppose that $I=(-\infty,\infty)$ and $p,r$ are periodic of period $\zeta$.
Define $P$ on $I$ as; 
                 $$ P(x)= \int_{0}^{x} p(t) dt , \forall x\in I    $$
Then the following statements are true


*

*A solution $\phi$ of the equation $y'+p(x)y=r(x)$ on $I$ is periodic of period $\zeta$ $\leftrightarrow$ $\phi(0)=\phi(\zeta)$.

*The equation $y'+py=r$ has exactly one periodic solution of  period $\zeta \leftrightarrow e^{P(\zeta)} \neq 1$.

*Every solution of the equation $y'+py=r$ is periodic of period $\zeta \leftrightarrow e^{P(\zeta)} =1$ AND $$ \int_{0}^{\zeta} e^{P(t)} r(t) dt=0 $$


I have already proved statement 1. I let $\phi$ be a solution of the equation $y'+p(x)y=r$.
Then I defined another function $\psi$ ...showed that $\psi$ is a solution of it also and after some computation solved that $\phi$ is periodic of period $\zeta$ using the fact that $\chi(x)= ce^{-P(x)}$.
Can anyone assist me in proving parts 2 and 3? Or atleast point me in the right direction. I'm really confused as to how to even begin these parts 
 A: For the sake of completeness, I'll start by tackling item (1) even though our OP Jason Moore expresses some satisfaction with his own solution.
1.)  Given that $\phi(x)$ is periodic of period $\zeta$, we have by definition that
$\phi(x + \zeta) = \phi(x), \; \forall x \in I = (-\infty, \infty); \tag 1$
thus if we take $x = 0$ we find that
$\phi(\zeta) = \phi(0). \tag{2}$
Next, we note that, if $\phi(x)$ satisfies
$y'(x) + p(x)y(x) = r(x), \tag 3$
that is,
$\phi'(x) + p(x)\phi(x) = r(x), \tag 4$
then we have
$\phi'(x + \zeta) + p(x + \zeta)\phi(x + \zeta) = r(x + \zeta),  \tag 5$
which, since $p(x)$ and $r(x)$ are periodic of period $\zeta$, that is, since
$p(x + \zeta) = p(x), \; r(x + \zeta) = r(x), \tag 6$
implies that
$\phi'(x + \zeta) + p(x)\phi(x + \zeta) = r(x); \tag 7$
we see that $\phi(x)$ and $\phi(x + \zeta)$ each satisfy (3)-(4), with initial conditions $\phi(0)$ and $\phi(0 + \zeta) = \phi(\zeta)$ respectively; since we assume that $\phi(\zeta) = \phi(0)$, $\phi(x)$ and $\phi(x + \zeta)$ both satisfy the same differential equation (3)-(4) with identical initial conditions $\phi(0) = \phi(\zeta)$, and thus by uniqueness of solutions we must have
$\phi(x + \zeta) = \phi(x),\; \forall x \in I; \tag 8$
the periodicity of $\phi(x)$ given $\phi(\zeta) = \phi(0)$ is thus established.
2.)  With
$P(x) = \displaystyle \int_0^x p(t)\; dt, \tag 9$
the solution $\phi(x)$ to (3)-(4) with initial condition $\phi(0)$ may be expressed in the form
$\phi(x) = e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt), \tag{10}$
which may easily be verified using direct differentiation; from (10),
$\phi'(x) = -P'(x)e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt) + e^{-P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt)'$
$= -P'(x)e^{P(x)}(\phi(0) + \displaystyle \int_0^x e^{P(t)}r(t) \; dt) + e^{-P(x)} e^{P(x)}r(x) = -p(x) \phi(x) + r(x), \tag{11}$
since, from (9),
$P'(x) = p(x); \tag{12}$
(11) is easily seen to be equivalent to (3)-(4); we also see that (10) implies
$\phi(0) = e^{-P(0)}(\phi(0) + \displaystyle \int_0^0 e^{P(t)}r(t) \; dt) = e^{P(0)}\phi(0) = \phi(0), \tag{13}$
since, again from (9),
$P(0) = \displaystyle \int_0^0 p(t) \; dt = 0; \tag{14}$
thus (10) is consistent with the initial conditions on (3)-(4) as well.
The formula (10) is in fact exceedingly well-known and its derivation my be found in many textbooks and at many web sites on the internet.
Now suppose
$e^{P(\zeta)} \ne 1; \tag{15}$
then
$e^{-P(\zeta)} \ne 1 \tag{16}$
as well, and if $x = \zeta$, (10) reads
$\phi(\zeta) = e^{-P(\zeta)}(\phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt), \tag{17}$
which may be re-written as
$e^{P(\zeta)} \phi(\zeta) - \phi(0) = \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt; \tag{18}$
we have seen in item (1) that $\phi(x)$ is a periodic solution with period $\zeta$ to (3)-(4) if and only if $\phi(\zeta) = \phi(0)$; if we choose $\phi(0)$, based upon (18), so that
$e^{P(\zeta)} \phi(0) - \phi(0) = (e^{P(\zeta)} - 1) \phi(0) = \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt, \tag{19}$
which is always possible by virtue of (15), then clearly (17) implies
$\phi(\zeta) = e^{-P(\zeta)}(\phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt) = \phi(0), \tag {20}$
and we may infer from item (1) that the trajectory initialized at $\phi(0)$ is periodic of period $\zeta$.  Furthermore, it is now clear from (19)-(20) that $\phi(0)$ is the only possible initial condition such that $\phi(\zeta) = \phi(0)$.
Thus we see that $e^{P(\zeta)} \ne  1$ implies the existence of a unique periodic solution to (2)-(3).
Now suppose there is precisely one periodic trajectory which has initial value $\phi(0)$, and that $e^{P(\zeta)} = 1$.  Then we also have $e^{-P(\zeta)} = 1$, and for this solution (20) must hold, whence
$\phi(\zeta) = \phi(0) + \displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt = \phi(0), \tag {21}$
which in turn implies
$\displaystyle \int_0^\zeta e^{P(t)}r(t) \; dt = 0; \tag{22}$
but, via (17) we find
$\phi(\zeta) = \phi(0), \tag{23}$
no matter what value $\phi(0)$ may take.  This shows that every solution must be periodic, which contradicts our assumption of a single periodic orbit.  Therefore we cannot have $e^{P(\zeta)} = 1$ if the periodic solution is unique; we conclude that $e^{P(\zeta)} \ne 1$ in this case.
Our demonstration of item (2) is thus complete.
3.)  We have already seen in item (2) that the combined requirements
$e^{P(\zeta)} = 1,\; \displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt = 0 \tag{24}$
are sufficient to imply every solution to (3)-(4) is periodic; so, supposing every $\phi(x)$ obeying (3)-(4) also satisfies
$\phi(\zeta) = \phi(0),\tag{25}$
then (20), and hence (19), holds for every value of $\phi(0)$; but as we have seen under item (2), $e^{P(\zeta)} \ne 1$ implies that (18) is valid for precisely one value of $\phi(0)$; we thus conclude that
$e^{P(\zeta)} = 1, \tag{26}$
but then (19) immediatly yields
$\displaystyle \int_0^\zeta e^{P(t)} r(t) \; dt  = 0, \tag{27}$
and we are done.
