What is the Laplace transform of $\dfrac{y'(t)}{t}$?

Question

I know that

${\mathcal{L}}\left\{ {\dfrac{1}{t}y\left( t \right)} \right\} = \int\limits_s^\infty {Y\left( u \right){\text{ d}}} u$

and that

${\mathcal{L}}\left\{ y'(t) \right\}=sY(s)-y(0)$

How can I find

${\mathcal{L}}\left\{ \dfrac{y'(t)}{t} \right\}$

Any help will be greatly appreciated

Thank you

@Babak : I think he just wants to see if there's a certain trick you can apply. I don't remember very well the properties of the Laplace transform, but feel free to assume anything you want on $f$, to me it looks like an exercise. — Patrick Da Silva, Aug 3, 2013 at 17:37
@user83382 : What happens if you try to compose the two given identities one after the other? Does something go wrong? — Patrick Da Silva, Aug 3, 2013 at 17:38
@PatrickDaSilva: Exactly, I was thinking about what you suggested. What goes wrong if we think of $\int_s^{\infty}\mathcal{L}(y'(t))du$? I thin the OP could do it as you commented. — Mikasa, Aug 3, 2013 at 18:09

Mikasa · Accepted Answer · 2013-08-03 18:26:45Z

3

As @Patrick suggested, I think the only way in which you can have a certain formula is to compose two formulas together. In fact:

$$\mathcal{L}\{y(t)/t\}=\int_s^{\infty} Y(u)du,~~Y(s)=\mathcal{L}\{y(t)\},~~\mathcal{L}\{y'(t)\}=sY(s)-y(0)$$ So:

$$\mathcal{L}\{y'(t)/t\}=\int_s^{\infty} \mathcal{L}\{y'(t)\}du=\int_s^{\infty} \big(uY(u)-y(0)\big)du$$

answered Aug 3, 2013 at 18:26

Mikasa

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$\begingroup$ Yes, I had something like that in mind, but I couldn't answer myself since I didn't want to bother remembering the details. +1 $\endgroup$
– Patrick Da Silva
Aug 3, 2013 at 19:19
$\begingroup$ The problem arose when I tried to apply Laplace transform to solve the following IVP $y'' + \dfrac{1}{t}y' + y = 0;\;y\left( 0 \right) = 1;y'\left( 0 \right) = 0$ but I realized that it isn't a good idea... $\endgroup$
– Raffaele
Aug 3, 2013 at 19:50
$\begingroup$ $\quad +^++^++^++^++\quad \ddot\smile$ $\endgroup$
– amWhy
Aug 4, 2013 at 1:21

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