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I'm having some trouble with this problem:

Solve the integral equation with Laplace transform

$$e^{-t}=y(t)+\int_0^t(t-u)y(u)du$$

I know how to use the Laplace transform for more "normal" equations but I don't understand this step here below in my solution manual. So they use the transform and get this

$$\frac{1}{s+1}=Y(s)+\frac{1}{s^2}Y(s)$$

can anyone derive or explain that last term for me? I have no idea how to get from that integral to that.

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    $\begingroup$ It's a known property of the Laplace transform that for the convolution of two functions $f$ and $g$, i.e. $f*g$, you have $\mathcal{L}\{f*g\}=F(s)G(s)$, where $F(s)$ and $G(s)$ are the respective transforms of $f$ and $g$. Are you wondering how this property arises? $\endgroup$ – user170231 Oct 14 '15 at 19:04
  • $\begingroup$ This I know, I just wonder where the $\frac{1}{s^2}$ factor came from $\endgroup$ – user269620 Oct 14 '15 at 19:23
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We have: $$L\left[\int_0^t(t-u)y(u)du\right]=Y(s)\cdot L[t]=Y(s)\cdot\frac{1}{s^2},$$ where in the first equality we used the convolution identity:

$$L\left[\int_0^t f(u)g(t-u)du\right]=F(s)\cdot G(s)$$

To clarify how we get the $\frac{1}{s^2}$: For $n\in\mathbb{N}$, we have:

$$L[t^n]=\frac{n!}{s^{n+1}}$$

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  • $\begingroup$ I have seen that general identity, its in my book but where does $\frac{1}{s^2}$ come from? I suspect that it's really easy but I just want to make sure that i get it. $\endgroup$ – user269620 Oct 14 '15 at 19:21
  • $\begingroup$ @Danny The Laplace transform of $t^n$ for $n$ a positive integer is $L[t^n]=\frac{n!}{s^{n+1}}$. $\endgroup$ – Ben Sheller Oct 14 '15 at 19:23
  • $\begingroup$ And $L[t]$ comes from the upper limit in the integral? $\endgroup$ – user269620 Oct 14 '15 at 19:25
  • $\begingroup$ @Danny $L[t]$ comes from using the function $g(t)=t$ in the convolution identity (so $t-u=g(t-u)$). $\endgroup$ – Ben Sheller Oct 14 '15 at 19:26
  • $\begingroup$ Aaaah okay, thanks a lot! $\endgroup$ – user269620 Oct 14 '15 at 19:28

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