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to get the inverse transform.) | to get the inverse transform.) | ||
| − | P.257 #8: Based on the other problems from this section, I | + | P.257 #8: Based on the other problems from this section, I factor out a t to get: f(t)=t^(n-1)e^kt but I don't really see how this helps us. Any thoughts on how to get this problem going would be appreciated. Thanks! |
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Revision as of 18:06, 18 October 2010
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)
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 difference of two angles and 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 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: Based on the other problems from this section, I factor out a t to get: f(t)=t^(n-1)e^kt but I don't really see how this helps us. Any thoughts on how to get this problem going would be appreciated. Thanks!
