Solving an integral equation by converting it to a differential equation

Question

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?!

Answer 1

\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.

Answer 2

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.

Answer 3

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?