Homework 7 collaboration area
Question: What exactly is 6.2, #9 asking when it says to use another method to find the laplace transform for Prob 1? (AM, 07-Oct)
Answer: I think they just want you to show that it can be computed in two ways. In problem 1, you probably used the identity
L[f'] = s F(s) - f(0).
To compute the same Laplace transform a second way, you could integrate directly from the definition of the Laplace transform, or maybe you could use
L[f"] = s^2 F(s) - s f(0) - f'(0)
to get the same answer as problem 1.
Sec6.2 P232 #31: I've factored out the s in the denominator so it looks like
$ \frac{1}{s}\ \frac{5}{s^2-5} $
But I'm not sure how to proceed from there.
Answer: You will need to use the integration formula on p. 239:
$ \mathcal{L}[\ \int_0^t f(\tau)d\tau \ ]=\frac{1}{s}F(s), $
using F(s) = 5/(s^2 - 5). Find f(t) and integrate as shown to find the inverse transform of the given function.
Sec6.3 P240 #8: I have it written out as
f(t)=[u(t-0)-u(t-pi)]*(1-e^(-t)).
I'm stuck on how to work out (1-e^(-t)). In the previous problem, #5, it was easy to make t^2 into
[(t-1)+1]^2 or [(t-2)+2]^2
and essentially not change the function. However, that's not the case with (1-e^(-t)) and I don't know what to do with it.
Answer: Do the same thing:
$ 1-e^{-t}=1-e^{-[(t-\pi)+\pi]}=1-e^{-\pi}e^{-(t-\pi)} $
Question: Pg. 232 - #9: The back of the book states that (cos a)^2 = 1/2 + 1/2 cos 2a.....where does this come from?
Answer: (cosx)^2=1-(sinx)^2, (sinx)^2=1/2-1/2cos(2x) => (cosx)^2=1-[1/2-1/2cos(2x)]=1/2+1/2cos(2x)
