Solving an integral equation by converting it to a differential equation I have to solve an integral equation given as
$$ x(y)=\sin(y)+\varepsilon \int_0^\infty e^{-s} x(y+s) \mathrm{d}s$$
I know that I have to differentiate it, but I am doing hard with it. Can anyone please help me?!
 A: A simple change of variables $t=s+y$ gives that
$$x(y)=\sin(y)+\varepsilon \int_y^\infty e^{-(t-y)} x(t) \mathrm{d}t=\sin(y)+\varepsilon e^y \int_y^\infty e^{-t} x(t) \mathrm{d}t.$$
Now differentiate it,
\begin{align*}
\dot{x}(y)&=\cos(y)+\varepsilon e^y \int_y^\infty e^{-t} x(t) \mathrm{d}t-\varepsilon e^ye^{-y}x(y)\\
&=\cos(y)+x(y)-\sin(y)-\varepsilon x(y)\\
&=(1-\varepsilon)x(y)+\cos(y)-\sin(y).
\end{align*}
Can you solve it now?
A: \begin{align}
 x(y)&=\sin(y)+\varepsilon \int_0^\infty e^{-s} x(y+s)\,ds\\
&=\sin(y)+\varepsilon  e^y\int^\infty_y e^{-s} x(s)\,ds
\end{align}
Hence
\begin{align}
e^{-y}x(y)=e^{-y}\sin(y)+\varepsilon\int^\infty_ye^{-s}x(s)\,ds
\end{align}
Setting $X(y)=\int^\infty_ye^{-s}x(s)\,dx$, we have thtat $X'(y)=-e^{-y}x(y)$. This gives  the first order differential equation
$$
\dot{X}+\varepsilon X=-e^{-y}\sin(y)
$$
the solution of this ODE is standard:
$$
(e^{\varepsilon y}X(y))'=e^{-(1-\varepsilon) y}\sin(y)
$$
integrating over $(0,y]$ on both sides, and performing integration by parts on the right-hand-side gives
$$
e^{\varepsilon y}X(y) = -\frac{1}{(1-\varepsilon)^2+1}e^{-(1-\varepsilon)y}\big((1-\varepsilon)\sin(y)+\cos(y)\big)+\frac{1}{(1-\varepsilon)^2+1}+X(0)
$$
Therefore
\begin{align}
 x(y)&=\sin(y)+\varepsilon e^{y}X(y)\\
&=\sin(y)-  \frac{\varepsilon}{(1-\varepsilon)^2+1}e^{-(1-\varepsilon)y}\big((1-\varepsilon)\sin(y)+\cos(y)\big)+\frac{\varepsilon}{(1-\varepsilon)^2+1}+\varepsilon X(0)
\end{align}
$X(0)=\int^\infty_0 e^{-s}x(s)\,dx$ is an initial condition of this problem.
A: You don’t have to differentiate in this case. The integrals are with respect to $s$, so you can simply factor out your $x$ and $y$ parts and evaluate them directly.
