Laplace Transform of equation

Question

I'm having trouble with the laplace transform: $\mathcal{L} \lbrace \sqrt{\frac{t}{\pi}}\cos(2 t) \rbrace$

The problem gives me the transform identity $\mathcal{L} \lbrace \frac{\cos(2 t)}{\sqrt{\pi t}} \rbrace = \frac{e^{-2/s}}{\sqrt{s}}$ but i'm not sure/confused as to why that would help me

Answer 1

$$ L[\sqrt{\frac{t}{\pi}}cos(2t)] $$

If you define the function: $f(t)=\sqrt{\frac{t}{\pi}}cos(2t)$

If you multiply it by $(\frac{t}{t})$ You can rewrite: $f(t)=t\frac{1}{\sqrt{ t\pi}}cos(2t)=t\frac{cos(2t)}{\sqrt{t\pi}}$

At this point, I have to remind you: " $L[t^nf(t)]=-(1)^n\frac{d}{ds^n}(L[f(t)])$ "

Finally, if you have $L[\frac{\cos(2 t)}{\sqrt{\pi t}}] = \frac{e^{-2/s}}{\sqrt{s}}$, you have to calculate $$ -(1)^n\frac{d}{ds}(\frac{e^{-2/s}}{\sqrt{s}}) $$

Answer 2

Assuming $t>0$ (which is a usual assumption with Laplace transforms), $$ \sqrt{\frac t\pi}\,\cos 2t=t\,\frac{\cos 2t}{\sqrt{\pi t}}. $$

Answer 3

HINT:

If $F(s) =\mathscr{L}\left(f(t)\right)(s)=\int_0^\infty f(t)e^{-st}\,dt$, then

$$-F'(s)=\int_0^\infty tf(t)e^{-st}\,dt$$

Now, let $f(t)=\frac{\cos (2t)}{\sqrt{\pi t}}$