Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

amberjordan2006 Group Title

.

  • one year ago
  • one year ago

  • This Question is Closed
  1. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmm it looks like we can do Partial Fraction Decomposition.

    • one year ago
  2. amberjordan2006 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And how would I do that?

    • one year ago
  3. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \frac{2s-3}{s^2-3s+2}\]The denominator can be factored, giving us,\[\large \frac{2s-3}{(s-2)(s-1)}\] From here, let's try applying Partial Fraction Decomposition. The factors in the denominator will break up like so, \[\large \frac{2s-3}{(s-2)(s-1)} \qquad = \qquad \frac{A}{s-2}+\frac{B}{s-1}\] 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,

    • one year ago
  4. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmm, 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.\[\large \cancel{(s-2)(s-1)}\frac{2s-3}{\cancel{(s-2)(s-1)}} \qquad = \qquad \left(\frac{A}{s-2}+\frac{B}{s-1}\right)(s-2)(s-1)\] And yes, after cancelling some stuff out on the right, you should get the thing you're thinking.

    • one year ago
  5. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hopefully you're coming up with something like this, \[\large 2s-3=A(s-1)+B(s-2)\] The s-2 's cancelled on the A term, while the s-1 's cancelled on the B term.

    • one year ago
  6. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    There 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.

    • one year ago
  7. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    A=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.

    • one year ago
  8. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Looks good c: Understand how to solve it from here?

    • one year ago
  9. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ummm, 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. \[\huge \mathscr{L}[e^{at}]=\frac{1}{s-a}\]Hopefully I'm remembering that correctly.

    • one year ago
  10. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    We'll want to be using it in this direction.\[\huge \mathscr{L}^{-1}\left[\frac{1}{s-a}\right]=e^{at}\]

    • one year ago
  11. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Here's just a quick reminder of where we've gotten so far, and the proper way to break it up. \[\large \mathscr{L}^{-1}\left[\frac{2s-3}{s^2-3s+2}\right]\qquad =\qquad \mathscr{L}^{-1}\left[\frac{1}{s-1}+\frac{1}{s-2}\right]\] \[\large =\qquad \mathscr{L}^{-1}\left[\frac{1}{s-1}\right]+\mathscr{L}^{-1}\left[\frac{1}{s-2}\right]\]

    • one year ago
  12. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmm 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.

    • one year ago
  13. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Cool 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

    • one year ago
  14. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes looks good! :) You have to use curly brackets on the exponent so it formats correctly hehe e^{2t}

    • one year ago
  15. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    If we were simply given F(s) at the start, not a differential equation, then I suppose we're done :O

    • one year ago
  16. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Yah we weren't given initial conditions or anything silly like that :) Luckily

    • one year ago
  17. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Close 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.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.