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= Homework 7 collaboration area = | = Homework 7 collaboration area = | ||
| − | 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) | + | 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. | ||
| + | |||
| + | Sec6.3 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: | ||
| + | |||
| + | <math>1-e^{-t}=1-e^{-[(t-\pi)+\pi]}=1-e^{-\pi}e^{-(t-\pi)}</math> | ||
| − | |||
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Revision as of 08:48, 8 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.
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)} $
