How we can show integral equation has a unique solution and how we can solve it? My question is on the title, e.g. how we can show that this integral equation has a unique solution and how we can estimate what the approximate answer near $0.1$ can be?
$$
x(t) + 0.1 \int_0^1 e^{ts}x(s) ds = 1, \quad (0 \leqslant t \leqslant 1).
$$
Thanks in advance.

What I've done so far:

By using the Binomial of $e^{ts}=(e^{t})^s$, we have:
$$
(e^s)^t = (\sum_{n=1}^{\infty}(\frac{s^n}{n!}))^t \approx (1+s)^t 
$$
then,
$$
(1+s)^t = \sum_{i=0}^t {t \choose i}s^i = 1 +ts+\frac{t(t-1)}{2!}s^2+\cdots +s^t \tag{$*$}
$$
by substituting $(*)$ in the integral we can estimate the solution, but i can not go any further.
 A: We use the Banach fixed-point theorem. First we recast the equation in a form that allows its application:
\begin{align}
x(t) = -0.1\int_{0}^{1} e^{ts}x(s) \, \mathrm{d}s + 1 =: Bx(t) + 1 =: Ax(t)
\end{align}
I'll assume $(C[0,1], \left\|\cdot\right\|_{\infty})$ is the Banach space of interest, but ostensibly you could use others ($L_1$ might be better). Now we attempt to prove $A$ is contractive on this space:
\begin{align}
\left\|Ax -Ay \right\|_{\infty} &= 0.1\sup_{t \in [0,1]}\left| \int_{0}^{1} e^{st}(x(s) - y(s)) \mathrm{d}s \right| \\
&\le 0.1\sup_{t \in [0,1]}e^{t} \int_{0}^{1} \left|x(s) - y(s)\right| \mathrm{d}s \\
&\le 0.1\left\|x-y\right\|_{\infty}
\end{align}
So $A$ is contractive with $\left\|B\right\|_{\infty} = 0.1$ and hence by the Banach fixed point theorem this equation has a unique solution.
Using the sequence $x_{n+1}:= Ax_{n}$, we have the estimate
\begin{align}
\left\|x_{n} - x\right\| \le \frac{0.1^{n}}{1-0.1}\left\|x_{1} - x_{0}\right\|_{\infty}.
\end{align}
Take $x_{0} = 0$, so that $x_{1} = 1$. Then
\begin{align}
x_{2}(t) = -0.1\int_{0}^{1}e^{st} \, \mathrm{d}s +1 = -\frac{0.1}{t}(e^{t} -1) + 1
\end{align}
This has the estimate
\begin{align}
\left\|x_{2} - x\right\|_{\infty} \le 0.011111
\end{align}
I'll leave it to you to calculate $x_{3}$.
