Homework 8 Collaboration Area
Question on problem 15 in Sec 6.6:
I tried to obtain the expression for
s/(s + 1) * 1/(s+1)
but am not getting the correct result in the Laplace table of
t sin t.
I am using the convolution of cos(tau)*sin(t-tau). There is no t term in sight. Is it okay to read off the table? Even if it is, shouldn't the result be the same?
Answer:
To find the inverse Laplace transform of
s/(s + 1) * 1/(s+1)
via convolution, you'll need to compute the convolution integral:
$ \int_0^t \cos(\tau)\sin(t-\tau)\ d\tau. $
You'll have to use a formula for the sine of the product of cosine and sine. Here it is:
(cos A)(sin B) = (1/2)sin(A+B) + (1/2)sin(B-A).
You'll need to be very careful. Remember, t acts like a constant in the integrals.
There is only one correct answer, so you should get it that way. (If it looks different than the back of the book, a trig identity might be at fault.)
Another, and perhaps better, way to solve this problem is to recognize that the given expression is the derivative of
1 / [(s+2)^2 + 1]]
...therefore greatly simplifying the solution (no trig identities required).
(You'll need to use the formula
L[ t f(t) ] = -F'(s)
to get the inverse transform.)
P.257 #8: I'm a bit confused on how to get this problem started... Based on other problems in this section, we can factor out a t such that the function inside the transform is t times f(t) where f(t)=t^(n-1)e^(kt), but I don't see how that helps us much.
Answer from Bell: You'll need to use mathematical induction for this problem. Start with n=1. Then use the formula
L[ t f(t) ] = -F'(s)
to reduce the n=2 case to the n=1 case, etc.
(AJ) Also, there's a better table of Laplace transforms at the end of chapter 6 that has the transform for f(t) = t^(n-1)e^(kt)
