Given $y''+y'-y=1$, how can I find :
All solutions for which $x \in [0,\infty)$
All periodic solutions
Since the solution is $y=c1*e^{(-\frac{1}{2} - \frac{\sqrt{5}}{2})x}+c2*e^{(-\frac{1}{2} + \frac{\sqrt{5}}{2})x} - 1$ -.Then for the third part $y(x)=y(x+T)$ for some $T>0$ which means that $y=c1*e^{(-\frac{1}{2} - \frac{\sqrt{5}}{2})x}*(1+e^{-T})+c2*e^{(-\frac{1}{2} + \frac{\sqrt{5}}{2})x}*(1+e^{-T}) =0$ ,then $1+e^-T$ = 0 or the period is $T=-\frac{1}{ln(e)}$
Solve a simpler problem first
$$ y''+y'- y=0 $$
instead of $1$.
What if you knew $y=C e^{rx}$ for some constants $C$ and $r$
\begin{eqnarray*} C r^2 e^{rx} + C r e^{rx} - C e^{rx} &=& 0\\ r^2 + r - 1 &=& 0\\ r &=& \frac{-1 \pm \sqrt{1+4}}{2}\\ &=& \frac{1}{2}(-1 \pm \sqrt{5}) \end{eqnarray*}
Call the $+$ solution $r_+$ and the minus solution $r_-$. So we know that the general solution to this simpler problem is a combination of $C_1 e^{r_+ x}$ and $C_2 e^{r_- x}$ where the constants $C_{1,2}$ haven't been solved for yet.
Say you had two solutions $y_p$ and $y_q$ of $y''+y'- y=1$. Then let $y_h=y_p-y_q$. You would see that
\begin{eqnarray*} y_h''+y_h'-y_h &=& (y_p'' + y_p - y_p) - (y_q''+y_q'-y_q)\\ &=& 1 - 1 = 0 \end{eqnarray*}
Because we have already solved $y''+y'-y=0$ we know that $y_h = C_1 e^{r_+ x} + C_2 e^{r_- x}$ for the $r_{\pm}$ we found above and some constants that we don't know $C_{1,2}$.
We only need to find one solution $y_p$ for the problem with $1$ on the RHS. Any other solution $y_q$ will just look like $y_p+y_h$ with appropriately chosen $C_{1,2}$. Much easier.
So we just have to find one solution $y_p$. Go with the simplest possible option. $y_p=-1$. The first two terms become $0$ on the LHS and we are left with $-(-1)=1$.