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working on a physical problem I arrived at the following equation $$ y(x) + A \int_{0}^{x} e^{\lambda (t-x)} y(t) \mathrm{d}t = 0$$ and after some struggling (not that easy to apply the basic Laplace transform tecnique) I had a look on EqWorld, where I discover that the equation $$y(x) + A \int_{0}^{x} e^{\lambda(x-t)}y(t)\mathrm{d}t = f(x)$$ has solution $$y(x) = f(x) - A \int_{0}^{x} e^{(\lambda-A)(x-t)} f(t)\mathrm{d}t$$ So, my equation being a particular case of the latter for f=0, I am lead to the conclusion only the nihil solution satisfies my equation, which is physically bizarre.Any hints here please?

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  • $\begingroup$ Is $\lambda$ a constant or a function? $\endgroup$ Nov 15 '13 at 18:10
  • $\begingroup$ Daniel, is a constant, my apologies $\endgroup$
    – Buco
    Nov 15 '13 at 18:26
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    $\begingroup$ No need for apologies. However, your integral equation yields $y(0) = 0$, and transforming it into a differential equation yields $y'(x) = -(\lambda+A)y(x)$, which has the unique solutions $y(x) = y(0) e^{-(\lambda+A)x}$. Why is it physically bizarre that $y\equiv 0$ is the only solution? $\endgroup$ Nov 15 '13 at 18:30
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    $\begingroup$ All correct. The equation arouse from considering the relaxation of a certain polymer upon the action of mechanical load and shrinking. Actually your comment made me think, the problem might be that, loosely speaking, the specimen has not been loaded! The value $y(0+)$ needs to be finite and positive, and then I would have exponentially decaying solutions asexpected, many thanks $\endgroup$
    – Buco
    Nov 15 '13 at 18:52
  • $\begingroup$ just a sign.. $y'-\lambda y+Ay=0$ so $y(x)=y(0)\operatorname{e}^{(\lambda-A)x}$ $\endgroup$
    – alexjo
    Nov 15 '13 at 21:50

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