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amberjordan2006
 2 years ago
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amberjordan2006
 2 years ago
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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm it looks like we can do Partial Fraction Decomposition.

amberjordan2006
 2 years ago
Best ResponseYou've already chosen the best response.0And how would I do that?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \frac{2s3}{s^23s+2}\]The denominator can be factored, giving us,\[\large \frac{2s3}{(s2)(s1)}\] From here, let's try applying Partial Fraction Decomposition. The factors in the denominator will break up like so, \[\large \frac{2s3}{(s2)(s1)} \qquad = \qquad \frac{A}{s2}+\frac{B}{s1}\] Where \(\large A\) and \(\large B\) are unknown constants that we need to solve for. The reason they're constants is because the unknown term you place on the top of each fraction, should be one degree lower than the denominator. Just something to maybe remember. :) Ok let's try solving this,

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm, yes it is :) It's good if you understand why you're doing that though. We'll start by multiplying through by the denominator on the left.\[\large \cancel{(s2)(s1)}\frac{2s3}{\cancel{(s2)(s1)}} \qquad = \qquad \left(\frac{A}{s2}+\frac{B}{s1}\right)(s2)(s1)\] And yes, after cancelling some stuff out on the right, you should get the thing you're thinking.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hopefully you're coming up with something like this, \[\large 2s3=A(s1)+B(s2)\] The s2 's cancelled on the A term, while the s1 's cancelled on the B term.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1There are several ways to continue from here. Since we have nice easy factors, the way I would recommend is by plugging in values for \(\large s\), (which is what I think you meant c; ). If you plug in \(\large s=1\) you should be able to find B very easily.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1A=1, B=1? Yah that's what I'm coming up with also. Remember the initial setup that we did? You want to plug the values back into that.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Looks good c: Understand how to solve it from here?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Ummm, I'm a little rusty on my Leplace Transforms. I think we just take the transform. This step requires a bit of memorization I guess. \[\huge \mathscr{L}[e^{at}]=\frac{1}{sa}\]Hopefully I'm remembering that correctly.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1We'll want to be using it in this direction.\[\huge \mathscr{L}^{1}\left[\frac{1}{sa}\right]=e^{at}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Here's just a quick reminder of where we've gotten so far, and the proper way to break it up. \[\large \mathscr{L}^{1}\left[\frac{2s3}{s^23s+2}\right]\qquad =\qquad \mathscr{L}^{1}\left[\frac{1}{s1}+\frac{1}{s2}\right]\] \[\large =\qquad \mathscr{L}^{1}\left[\frac{1}{s1}\right]+\mathscr{L}^{1}\left[\frac{1}{s2}\right]\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm I think you'll get two terms out of this one. When your \(\large a\) is \(\large 1\), it should give you \(\large e^{t}\). Which is what that first inverse leplace should be giving us. It looks like you have the second term correctly figured out though.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Cool looks like you were able to get through this one without too much trouble \c:/ Just needed to brush up on those darn Partial Fractions! heh

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yes looks good! :) You have to use curly brackets on the exponent so it formats correctly hehe e^{2t}

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1If we were simply given F(s) at the start, not a differential equation, then I suppose we're done :O

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Yah we weren't given initial conditions or anything silly like that :) Luckily

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Close this thread at the top, it's getting too long. It's going to get laggy with all of the fancy latex formatted. Open a new thread with your next question. I'll look for it. You can type @zepdrix in the comments somewhere if I'm taking too long, heh. That will send me a page and I can find yer question easier.
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