Solving a non-homogeneous Volterra integral equation of the second kind 
Q. If $y(x)=1+\displaystyle\int_0^x e^{-(x+t)}y(t)\,dt,$ then $y(1)$ equals:
(a) $0$, (b) $1$, (c) $2$, (d) $3$.

I tried the successive approximations method (starting with $y_0=0$), conversion to a DE, and finding the resolvent kernel. But none of them worked out for me! So, I thought of applying the Laplace transform technique:
\begin{align}
\mathscr{L}[y(x)]&=\mathscr{L}[1] + \mathscr{L}\left[\int_0^x e^{-(x+t)}y(t)\,dt\right]\\
&=\frac 1s + \frac 1s \mathscr{L}[e^{-(x+t)}y(t)]
\end{align}
In $\mathscr{L}[e^{-(x+t)}y(t)]$, both $x$ and $t$ are variables. Is it even possible to find $\mathscr{L}[e^{-(x+t)}y(t)]$ ?
 A: Let us rewrite the integral as follows:
\begin{align*} 
\int_0^x e^{-(x + t)}y(t)\, dt = e^{-2x}\int_0^x e^{x - t}y(t)\, dt = e^{-2x}f(x). 
\end{align*}
Let $Y(s) = \mathcal{L}\{y(x)\}$. Set $g(x) = e^x$. The convolution property gives
\begin{align*} 
F(s) = \mathcal{L}\{f(x)\} = \mathcal{L}\left\{\int_0^x g(x - t)y(t)\, dt\right\} = G(s)Y(s) = \frac{1}{s - 1}Y(s). 
\end{align*}
The translation property then yields
\begin{align*} 
\mathcal{L}\left\{e^{-2x}f(x)\right\} = F(s + 2) = \frac{1}{s + 1}Y(s + 2). 
\end{align*}
Transforming the given integral equation, we get
\begin{align*} 
Y(s) & = \frac{1}{s} + \frac{1}{s + 1}Y(s + 2)\\ 
Y_{s + 2} - (s + 1)Y_s & = -\frac{s + 1}{s} = -1 - \frac{1}{s}, 
\end{align*}
which is a second-order nonhomogeneous difference equation.
A: Observe that $y\ge 0.$ Indeed we have $y(0)=1.$ Let $x_0$ be the smallest point such that $y(x_0)=0.$ Then $y(t)>0$ for $0\le t<x_0$ and
$$0=y(x_0)=1+\int\limits_0^{x_0}e^{-(x_0+t)}y(t)\,dt >1$$
a contradiction. The positivity of $y$ implies
$y(x)>1$ for $x>0.$ Next we will show that $$y(x)<{4\over 3},\quad x\ge 0\quad (*)$$
Assume for a contradiction that $(*)$ fails. Let $x_1$ be the smallest point in $[0,1]$ such that $y(x_1)={4\over 3}.$ Then $x_1>0,$ $y(t)<{4\over 3}$ for $0\le t\le x_1$ and
$${4\over 3}=y(x_1)=1+\int\limits_0^{x_1}e^{-(x_1+t)}y(t)\,dt\le 1+{4\over 3}e^{-x_1}\int\limits_0^{x_1}e^{-t}\,dt\\
= 1+{4\over 3}e^{-x_1}(1-e^{-x_1})={4\over 3}-{1\over 3}\left (2e^{-x_1}-1\right )^2< {4\over 3}$$ a contradiction.
Summarizing we have obtained $1<y(x)<{4\over 3}$ for any $x>0.$ In particular there holds $1<y(1)<{4\over 3}.$
A: I am not exactly sure about the approach using Laplace Transform. However, I can suggest the following method for solving this:
First, differentiate both sides of the equation using Leibnitz rule, and note that $y(0)=1$.
So we get
$$y'(x)=x\dfrac{d}{dt}\left[e^{-(x+t)}y(t)\right]|_{t=x}=x\left[-e^{-(x+x)}y(x)+e^{-(x+x)}y'(x)\right]$$
$$\Rightarrow y'(x)= xe^{-2x}\left[y'(x)-y(x)\right]$$
$$\Rightarrow (1-xe^{-2x})y'(x)= -xe^{-2x}y(x)$$
$$\Rightarrow \dfrac{d\left(\ln y\right)}{dx}= -\dfrac{xe^{-2x}}{1-xe^{-2x}}$$
This gives $$y = \exp\left[ \int_0^x \dfrac{te^{-2t}dt}{te^{-2t}-1}\right]$$
Hope this helps.
A: $$y(x)=1+\displaystyle \int_0^x e^{-(x+t)}y(t)\,dt, \implies y(x)=1+e^{-x}\int_{0}^{x} e^{-t} y(t) dt $$
So $y(0)=1$,D.w.t.$x$ using Lebnitz as
$$y'(x)=-e^{-x}\int_{0}^{x} e^{-t} y(t) dt+e^{-x} e^{-x} y(x).$$
$$\implies y'(x)=-(y-1)+e^{-2x} y(x)$$
$$\implies y'(x)+(1-e^{-2x})y(x)=1$$
This is liner ODE whose integrating factor is $$I=e^{\int (x-e^{-2x})dx}=e^{x+\frac{1}{2}e^{-2x}}.$$
So,
$$y(x)=e^{-x-\frac{1}{2}e^{-2x}}\int e^{x+\frac{1}{2}e^{-2x}} dx+ C e^{-x-\frac{1}{2}e^{-2x}}$$
$$y(x)=1-\sqrt{\frac{\pi}{2}}e^{-x-\frac{1}{2}e^{-2x}} \text{Erfi}(e^{-x}/\sqrt{2})+C e^{-x-\frac{1}{2}e^{-2x}}$$
Note that $\int e^{x+\frac{1}{2}e^{-2x}} dx=e^{x+\frac{1}{2}e^{-2x}}-\sqrt{\frac{\pi}{2}} \text{Erfi}(e^{-x}/\sqrt{2}).$
