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 $\mathscr{L}[e^{-(x+t)}y(t)]$ ?

$$y(x)=1+\displaystyle \int_0^x e^{-(x+t)}y(t)\,dt, \implies y(x)=1+e^{-x}\int_{0}^{x} e^{-t} y(t) dt $$
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 $\int e^{x+\frac{1}{2}e^{-2x}} dx=e^{x+\frac{1}{2}e^{-2x}}-\sqrt{\frac{\pi}{2}} \text{Erfi}(e^{-x}/\sqrt{2}).$
