So I have a problem (two problems, actually) that a friend helped me out with, I'm able to work out the components of this problem but get lost when I have to bring it all together... so what I have is this
$f(t)=t^{2}e^{-2t}+e^{-t}\cos(3t)+5$
Simple enough. I got:
$$\mathcal{L}[t^2]=\frac{2}{s^2}$$
$$\mathcal{L}[e^{-2t}]=\frac{1}{s+2}$$
$$\mathcal{L}[e^{-t}]=\frac{1}{s+1}$$
$$\mathcal{L}[\cos(3t)]=\frac{s}{s^2+9}$$
$$\mathcal{L}[5]=\frac{5}{s}$$
Now the problem is bringing it all together... Apparently,
$$\mathcal{L}[t^2e^{-2t}]=\frac{2}{(s+2)^2}$$
And
$$\mathcal{L}[e^{-t}\cos(3t)]=\frac{s}{(s+1)^2+9}$$
and I just don't seem to get why...? how do you bring those guys together like that? whats the reason behind it?
the other problem is much simpler and I'm sure I have the correct answer... it is:
$$t^3-5\cos(5t)$$
I get:
$$\mathcal{L}[t^3]=\frac{6}{s^4}$$
And
$$\mathcal{L}[5\cos(5t)]=\frac{s}{s^2+25}$$
which together is supposedly,
$$\mathcal{L}[t^3-5\cos(5t)]=\frac{6}{s^4}-\frac{5s}{s^2-25}$$
Is it correct? Any help at all is much appreciated. Thank you.
No sorry. Your answer for one part is wrong. We have: $$L\left[e^{-t} \cos (3t)\right]=\frac{s+1}{(s+1)^2+9}$$
Also, $$L \left[5\cos (5t)\right] = \frac{5s}{s^2+25}$$ a typo perhaps?
Note that if $f(t) $ has the transform $F(s) $, then $$L\left[e^{at} f(t) \right] = F(s-a) $$ This is called the First Shifting Theorem.