Revision as of 09:59, 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.

Sec6.2 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:

$1-e^{-t}=1-e^{-[(t-\pi)+\pi]}=1-e^{-\pi}e^{-(t-\pi)}$

@@ Line 34: / Line 34: @@
 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.
+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: