Consider the integral equation $\phi(x)-\frac{e}{2} \int_{-1}^{1} x e^{t} \phi(t) d t=f(x) .$ Then
- there exists a continuous function $f:[-1,1] \rightarrow(0, \infty)$ for which solution exists
- there exists a continuous function $ f:[-1,1] \rightarrow(-\infty, 0)$ for which solution exists
- for $f(x)=e^{-x}\left(1-3 x^{2}\right)$, a solution exists
- for $f(x)=e^{-x}\left(x+x^{3}+x^{5}\right),$ a solution exists
what I tried
$f(x)=f(x)+\frac{e}{2} \int_{-1}^{1} x e^{t} \phi(t) d t$
$\phi(x)=f(x)+\frac{e x}{2} \int_{-1}^{1} e^{t} \phi(t) d t$
$ \begin{array}{l}L e t \quad c=\int_{-1}^{1} e^{t} \phi(t) d t \\ \Rightarrow \phi(x)=f(x)+\frac{x e}{2} c\end{array} $
$c=\int_{-1}^{1} e^{t}\left(f(t)+\frac{c e}{2} t\right) d t$
$c=\int_{-1}^{1} e^{t} f(t)+\frac{\operatorname{ce}}{2} \int_{-1}^{1} t e^{t} d t$
$c=\int_{-1}^{1} e^{t}f( t)+\frac{\operatorname{ce}}{2}\left[t e^{t}-e^{t}\right]_{-1}^1$
$c=\int_{-1}^{1} e^{t} f(t)+\frac{c e}{2}\left(2 e^{-1}\right)$
$c=\int_{-1}^{1} e^{t} f(t) d t+c$
$\int_{-1}^{1} e^{t} f(t) d t=0$ Option (3) and (4) satisfies this Relation so, (3) and (4) seems true to me.is i am right?
I am stuck for option (1) and (2) is there any result or theorem ?
If $f(t)$ is completely positive or negative according to options $1$ and $2$ respectively, $\int_{-1}^{1} e^{t} f(t) d t=0$ would not hold, for $e^t>0\ \forall\ t\in(-\infty,+\infty)$! An obvious guess for $f(t)$ is $g(t)\times e^t$, where $g(t)$ is an odd function, since $\int_{-1}^{1} g(t)dt=0$. Other cases, like the one you have mentioned $e^xf(x)=1-3x^2$, also fit well when they at least change sign for $x\in[-1,+1]$.