Differential equations help please

**DE novice** · July 17th 2008, 09:14 AM

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 ?

**mr fantastic** · July 17th 2008, 03:14 PM

Originally Posted by DE novice

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 ?

For the benefit of others who will look at this thread, the questions are:

1. Solve:

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

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

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

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

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.

**topsquark** · July 17th 2008, 05:14 PM

Is that $\mu _2$ in 1a) supposed to be a unit step function?

-Dan

**DE novice** · July 18th 2008, 07:46 AM

Originally Posted by topsquark

Is that $\mu _2$ in 1a) supposed to be a unit step function?

-Dan

yes. From what i understand it means the function mu equals 0 when t < 2 and 1 when t >= 2

**DE novice** · July 19th 2008, 12:57 PM

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.

**mr fantastic** · July 20th 2008, 07:18 PM

Originally Posted by DE 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.

(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

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

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

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

By the way, don't bump. Replies are given when people have time.

**DE novice** · July 21st 2008, 07:46 AM

Originally Posted by mr 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

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

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

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

By the way, don't bump. Replies are given when people have time.

Sorry for the bump. My apologies

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

as far as 2a, yes the theorem $LT^{-1} \left[e^{-as} F(s)\right] = f(t - a) u(t - a)$ 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 ?

**mr fantastic** · July 21st 2008, 07:07 PM

Originally Posted by DE novice

Sorry for the bump. My apologies

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

as far as 2a, yes the theorem $LT^{-1} \left[e^{-as} F(s)\right] = f(t - a) u(t - a)$ 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 ?

Not quite (to me, anyway).

I'd present the final lines something like this:

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

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

Therefore $y = \, .......$

**DE novice** · July 23rd 2008, 08:23 AM

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.

**mr fantastic** · July 23rd 2008, 05:06 PM

Originally Posted by DE 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.

Looks OK.

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