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


1 Answer 1


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,


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 $x=0$:



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