Existence of a unique solution. Consider the integral equation for some given function $y(t)\in C[0,1]$ and a given constant $\lambda$ with $|\lambda|<1$,
$$x(t)-\lambda\int_0^1 e^{t-s}x(s)ds=y(t).$$
Show that there exists a unique solution $x(t)\in C[0,1]$.
 A: First, set $\displaystyle x_0(t) = y(t) + \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s$. Because$$\begin{align*}
y(t) + λ \int_0^1 \mathrm{e}^{t - s} x_0(s) \,\mathrm{d}s &= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + λ \int_0^1 \mathrm{e}^{t - s} \left(\frac{λ}{1 - λ}\int_0^1 \mathrm{e}^{s- u} y(u) \,\mathrm{d}u\right) \,\mathrm{d}s\\
&= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u\mathrm{d}s\\
&= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u \int_0^1 \mathrm{d}s\\
&= y(t) + λ \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s + \frac{λ^2}{1 - λ} \int_0^1 \mathrm{e}^{t - u} y(u) \,\mathrm{d}u\\
&= y(t) + \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{t - s} y(s) \,\mathrm{d}s = x_0(t),
\end{align*}$$
then $x_0(t)$ is a solution to the given equation.
Now suppose $x(t)$ is a solution to the given equation, then$$
x(t) = y(t) + λ \int_0^1 \mathrm{e}^{t - s} x(s) \,\mathrm{d}s = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s.
$$
Denote $\displaystyle c = λ \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s$, then $x(t) = y(t) + c\mathrm{e}^t$. Therefore,$$\begin{align*}
y(t) + c\mathrm{e}^t &= x(t) = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} x(s) \,\mathrm{d}s = y(t) + λ \mathrm{e}^t \int_0^1 \mathrm{e}^{-s} (y(s) + c\mathrm{e}^s) \,\mathrm{d}s\\
&= y(t) + λ \mathrm{e}^t \left(\int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s + c\right),
\end{align*}$$
which implies$$
c = λ \left(\int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s + c\right),
$$
i.e.$$
c = \frac{λ}{1 - λ} \int_0^1 \mathrm{e}^{-s} y(s) \,\mathrm{d}s.
$$
Hence $x(t) = y(t) + c\mathrm{e}^t \equiv x_0(t)$, so the solution to the given equation is unique.
