How to show $e^t$ is the unique solution to the integral equation $\int_0^1 e^{ts} x(s) ds = \frac{e^{t+1}-1}{t+1}$? How to show $x(t) = e^t$ is the unique solution to the following integral equation? $$\int_0^1 e^{ts} x(s) ds = \frac{e^{t+1}-1}{t+1}$$
My thoughts:
$$(t+1)\int_0^1 e^{ts} x(s) ds = e^{t+1}-1 $$
Integrating by parts gives:
$$\int_0^1 e^{ts}(x(s)-x'(s)) ds = (e - x(1))e^{t} + x(0) - 1$$
Taking the $k$-th derivative of both sides about $t$:
$$\int_0^1 s^k e^{ts}(x(s)-x'(s)) ds = (e - x(1))e^{t}$$
Let $t = 0$
$$\int_0^1 s^k (x(s)-x'(s)) ds = e - x(1)$$
I cannot figure out how to proceed next.

Suppose $x(s)$ and $y(s)$ both satisfy the integral equation. Then  $\int_0^1 e^{ts}(x(s)-y(s))ds=0$ for all $t$. Then the same is true if we replace $e^{ts}$ with a finite linear combination of elements of the family
 $\left\{e^{ts}\right\}_{t\in \mathbb{R}}$. Thus 
$$\int_0^1 f(s)(x(s)-y(s))ds,\qquad \forall f\in \operatorname{span}\left\{e^{ts}\right\}_{t\in \mathbb{R}} $$ 
But $\operatorname{span}\left\{e^{ts}\right\}_{t\in \mathbb{R}}$ is a subalgebra of $C([0,1])$ which separates points and contains a nonzero constant function (let $t=0$), and therefore by the Stone-Weierstrass theorem, it is dense in $C([0,1])$. Thus, using a standard limiting under the integral sign procedure,
$$\int_0^1 f(s)(x(s)-y(s))ds=0,\qquad \forall f\in C([0,1]) $$
It is now a known fact in analysis that this implies $x(s)-y(s)\equiv 0$, i.e. $x=y$. Thus the solution to your integral equation is unique.
