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I have a question about my try to solve the following equation: $y(x)=2\cos(x)+\epsilon\int_0^\infty (\frac{1}e)^\tau*y(x+\tau) d\tau $ where y is bounded.

My approach was write out the improper integral $y(x)=2cos(x)+\epsilon \lim_{a\to \infty}\int_0^a (\frac{1}e)^\tau*y(x+\tau) d\tau$

and than to differentiate the equation, so that i get $y'(x)=-2sin(x)+\epsilon \lim_{a\to \infty}(\frac{1}e)^a*y(x+a)$

since y is bounded there exists A in $\mathbb{R}$, such that y(x)$\leq$ A for all x. With this in mind i get $y'(x)=-2sin(x)+\epsilon \lim_{a\to \infty}(\frac{1}e)^a*y(x+a)$ =-2sin(x)

(After that one just needs to solve y(x)=-2sin(x))
Now i am not really sure if this approach is right and would like to ask you for your feedback.

thanks in advance

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Differentiating both sides of the integral equation with respect to $x$, we obtain

$$y'(x)=-2\sin(x)+\epsilon\int_0^\infty e^{-\tau}\partial_{x}y(x+\tau) d\tau.$$

Since $\partial_x y(x+\tau)=\partial_{\tau} y(x+\tau)$, integration by parts yields

\begin{align*} y'(x)&=-2\sin(x)+\epsilon\int_0^\infty e^{-\tau}\partial_{\tau}y(x+\tau) d\tau \\ &= -2\sin(x)+\epsilon\left[e^{-\tau}y(x+\tau)|_0^{\infty}+\int_0^{\infty}e^{-\tau}y(x+\tau) d\tau \right] \\ &= -2\sin(x)-\epsilon y(x)+y(x)-2\cos(x), \end{align*}

where the last equality follows from the initial integral equation. Thus, we have reduced the problem to the solution of an inhomogeneous linear differential equation,

$$y'(x)-(1-\epsilon)y(x)=-2\sin(x)-2\cos(x),$$

which can be solved using well known methods. Being a first order ODE, its general solution $\tilde{y}(x)$ has an arbitrary integration constant, which can be determined by plugging $\tilde{y}(x)$ in the original integral equation and solving it for $x=0$:

$$\tilde{y}(0)=2+\epsilon\int_0^{\infty}e^{-\tau}\tilde{y}(\tau)d\tau.$$

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