2
$\begingroup$

If, $$\mathcal L \left\{ \frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}} \right\}= \frac{\exp\big(\frac{-3}{s}\big)}{\sqrt{s}}$$, $$\mathcal L^{-1} \left\{ \frac{\exp\big(\frac{-1}{s}\big)}{\sqrt{s}}\right\}=?$$

could help

Laplace transform, proof that $L \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$

Thanks!

$\endgroup$
6
  • $\begingroup$ I think it might be useful math.stackexchange.com/questions/215831/… $\endgroup$
    – P. M. O.
    Oct 17, 2012 at 19:49
  • $\begingroup$ Do you know the answer? $\endgroup$
    – vesszabo
    Oct 17, 2012 at 19:50
  • $\begingroup$ @vesszabo No... $\endgroup$
    – P. M. O.
    Oct 17, 2012 at 19:51
  • $\begingroup$ Why do you think Laplace transform, proof that ... could help? $\endgroup$
    – vesszabo
    Oct 17, 2012 at 19:53
  • $\begingroup$ @P.M.O.: I think there is a typo in the body of your question. $\endgroup$
    – Mikasa
    Oct 17, 2012 at 20:21

2 Answers 2

3
$\begingroup$

You noted that: $$\mathcal L \{ \frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}} \}= \frac{\exp(\frac{-3}{s})}{\sqrt{s}}$$ and $$\mathcal L \{ \frac{1}{k}f\bigg(\frac{t}{k}\bigg) \}= F(ks)$$ and $F(s)=\frac{\exp(\frac{-3}{s})}{\sqrt{s}}$. Set $s$ to $3s$ in $F(s)$, so you have $$\sqrt{3}F(3s)=\frac{\exp(\frac{-1}{s})}{\sqrt{s}}$$ So $$\mathcal L^{-1}\{\sqrt{3}F(3s)\}=\mathcal L^{-1} \bigg(\frac{\exp(\frac{-1}{s})}{\sqrt{s}}\bigg)$$ But $$\mathcal L^{-1}\{\sqrt{3}F(3s)\}=\frac{1}{\sqrt{3}}\frac{\cos(2\sqrt{3t})}{\sqrt{\pi t}}\bigg|_{t\to t/3}=\frac{\cos(2\sqrt{t})}{\sqrt{\pi t}}$$

$\endgroup$
1
$\begingroup$

A slight modification of the method you used to compute the first transform will give you:

$$\mathcal{L}\left(\frac{\cos\left(k\sqrt{t}\right)}{\sqrt{\pi t}}\right) = \frac{e^{-k^2/4s}}{\sqrt{s}} \, . $$

For your inverse transform you need $k=2$.

$\endgroup$
1
  • $\begingroup$ how i can use $\mathcal{L} \{ \frac{1}{k}f(\frac{t}{k}) \}= F(ks)$ $\endgroup$
    – P. M. O.
    Oct 17, 2012 at 20:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .