Help ! I'm working Laplace Tranforms with periodic properties & convolution and I'm not getting it. Can someone please show me how to solve these 2 problems (attached) step by step so i can understand ?

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- Jul 17th 2008, 09:14 AMDE noviceDifferential equations help please
Help ! I'm working Laplace Tranforms with periodic properties & convolution and I'm not getting it. Can someone please show me how to solve these 2 problems (attached) step by step so i can understand ?

- Jul 17th 2008, 03:14 PMmr fantastic
For the benefit of others who will look at this thread, the questions are:

1. Solve:

(a) $\displaystyle y'' + 3y' + 2y = \mu_2 (t), ~ y(0) = 0, ~ y'(0) = 1$.

(b) $\displaystyle y'' - y' + 2y = 20 \, \delta (t-3), ~ y(0) = 1, ~ y'(0) = 0$.

2. Solve (express the answer as a convolution):

$\displaystyle 4y'' + 4y' + 17y = g(t), ~ y(0) = 0, ~ y'(0) = 1$.

As a matter of courtesy you should have done what I've done above, especially given the sort of reply you're hoping to get.

There is a lot of*routine*hack work that can be done simply by applying the various formulae. So what I want to know is what have*you*done so far? Where exactly are you stuck?

Please show all the work you've been able to do.

You will also find in these forums (use the search option) several detailed solutions to similar questions. - Jul 17th 2008, 05:14 PMtopsquark
Is that $\displaystyle \mu _2$ in 1a) supposed to be a unit step function?

-Dan - Jul 18th 2008, 07:46 AMDE novice
- Jul 19th 2008, 12:57 PMDE novice
Here's what I have for #1a and 1b (I have it worked in the attachment as #2a & b but that's just how its numbered in my homework).....sorry its an attachment but it would be alot of stuff to key in and I'm not sure how to make it view in the message text ..still working on #2. Comments welcomed. Thanks.

- Jul 20th 2008, 07:18 PMmr fantastic
(2a) looks fine, but I don't see how your answer (which looks correct) follows from the third and second last lines ...... Surely you'd just apply the theorem

$\displaystyle LT^{-1} \left[e^{-as} F(s)\right] = f(t - a) u(t - a)$

at the fourth last line ........

As for (2b), won't the second term of the answer contain $\displaystyle \sinh (t - 3)$ rather than $\displaystyle \cosh (t - 3)$ .......?

By the way, don't bump. Replies are given when people have time. - Jul 21st 2008, 07:46 AMDE novice
Sorry for the bump. My apologies

Thank you so much for your good catch on 2b !!!!!

as far as 2a, yes the theorem $\displaystyle LT^{-1} \left[e^{-as} F(s)\right] = f(t - a) u(t - a)$ was what I was trying to apply but I just wanted to make sure I was using it correctly. Does it look like that is the case ? - Jul 21st 2008, 07:07 PMmr fantastic
Not quite (to me, anyway).

I'd present the final lines something like this:

But $\displaystyle LT^{-1} \left[ e^{-2s} \left( \frac{1}{2} \, \frac{1}{s} - \frac{1}{s+1} + \frac{1}{2} \, \frac{1}{(s+2)} \right) \right] = u(t-2) \, f(t - 2)$

where $\displaystyle f(t) = LT^{-1} \left[ \frac{1}{2} \, \frac{1}{s} - \frac{1}{s+1} + \frac{1}{2} \, \frac{1}{(s+2)} \right] = \, ...... $

Therefore $\displaystyle y = \, ....... $ - Jul 23rd 2008, 08:23 AMDE novice
Here's what I have for problem #2 listed in the original post to this thread (I have it ID'd as #3 in the attachment b/c that's what # it is in my homework). Could someone please verify my answer ? Thx.

- Jul 23rd 2008, 05:06 PMmr fantastic