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I don't know how to go about solving these questions the second one seems to want integration by parts but I don't know how that will work out, an the first one seems to need the definition of Laplace transforms but I don't see how that will work out.

Show that for any function f(t) for which the Laplace transform $L [f(t)] = F(s)$ exists, $L[tf(t)] = −F'(s)$

Solve the equation for $x(t)$ : $$\int_0^{t}(x(τ )(t− τ ))dτ = t^4$$

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  • $\begingroup$ So your equation looks like a convolution $x \ast h(t)$ with $h(t) = t$ and $x(t)$ causal. But because of the non-causality of $h(t)$, for solving it I need the Fourier transform (of distributions... more or less the bilateral Laplace transform) . Are you sure it is $\int_0^\infty$ and not $\int_0^t$ ??? $\endgroup$
    – reuns
    Jul 12, 2016 at 9:41
  • $\begingroup$ Your correct thanks. $\endgroup$
    – user353452
    Jul 12, 2016 at 9:45
  • $\begingroup$ So if $y(t) = x \ast h(t)$ then $Y(s) = \ ?$ (here $h(t) = t \ 1_{t > 0}$) $\endgroup$
    – reuns
    Jul 12, 2016 at 9:53
  • $\begingroup$ Wait what I'm confused. $\endgroup$
    – user353452
    Jul 12, 2016 at 10:00
  • $\begingroup$ $y(t) =\int_0^{t}x(\tau )(t− \tau)d\tau = x \ast h(t)$ (where $\ast $ is the convolution), $Y(s) = \int_0^\infty y(t) e^{-st} dt = \ ?$ $\endgroup$
    – reuns
    Jul 12, 2016 at 10:02

1 Answer 1

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You can solve this problem by using the property of convolution theorem.

Following are the steps to solve these type of problems:

First, take the Laplace transform on both the sides.
Then, on the left hand side apply convolution theorem in which $f(t)=x(t)$ and $g (t) = t$. After calculating $x(s)$, take the inverse transform.

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