Revision as of 11:39, 11 October 2010

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.

For 6.2, #9 (prob 1 part) I followed the book's hint and used shifting. See Theorem 2 pp. 224. Let f(t)=t, and use $ e^{kt} $ for shifting. --Rekblad 06:57, 11 October 2010 (UTC)

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.

Question continued: I understand the above formula, but why is F(s) = 5/(s^2 - 5) if the quantity we are trying to evaluate is L^-1[5/(s^3 + 5)]? I see that you can factor out the 1/s, but I still would think F(s) = 5/(s^3 + 5)...

And, if you are able to prove to me that I'm just being an idiot, maybe you can do it again: How do you find the inverse Laplace of (1/s)*(5/(s^2 + 5))? The first part is just 1 and the second is sqrt(5)*sin(sqrt(5)t) but I doubt that's the correct way to go about this.

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)} $

Sec6.3 #5: Yes, I agree it is easy to make t^2 into [(t-1)+1]^2 or [(t-2)+2]^2. Doing this, the book's solution is 2/s^3 + 2/s^2 + 1/s but I get: ... - 1/s by expanding upon the previous square. Any thoughts as to why the sign difference? Thanks.

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)

Question: pg247 #7: Does anyone think that there might be a typo in the back of the book? in the "u(t-4)" term, I don't see how they have 1/3sin(3t-12), I think it should be 2/3. The 1/3e^-t*sin3t makes sense because you have a (1-2/3)e^-t*sin33t, but there are no other sine terms to combine with in the "u(t-4" area. This problem was heavy on the book keeping and I might have dropped something, but I can't find it.

Answer: I actually made the same mistake my first go around. The problem I encountered was after solving for the partial fractions of the "u(t-4)" term, I had:

$ \frac{-1}{s}\ + \frac{s+2}{(s+1)^2+9}\ $

This should expand to:

$ \frac{-1}{s}\ + \frac{s+1}{(s+1)^2+9}\ + \frac{1}{(s+1)^2+9}\ $

This will give you the answer the book has. Our error came from violating the s-shifting rules and incorrectly solving for the inverse Laplace using the following:

$ \frac{-1}{s}\ + \frac{s}{(s+1)^2+9}\ + \frac{2}{(s+1)^2+9}\ $

As you can see, the inverse laplace cannot be taken of the second term.... Hope this helps you.

It does, I can't believe we both made the same mistake. Thank you.

Page 240, Problem 16. I don't understand what to do with the e^-s term. Any guidance on this problem?

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