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LaPlace D.E. Question #9

  1. Apr 15, 2013 #1
    y"-8y'+20y=tet, y(0)=0, y'(0)=0

    I need to know if I made a mistake in getting to the step below:

    L-1{ 1/[(s-1)2(s2-8s+20)] }

    because when I solve that, I get something pretty gnarly..
     
  2. jcsd
  3. Apr 15, 2013 #2

    Mark44

    Staff: Mentor

    Show us how you got to that step.
     
  4. Apr 15, 2013 #3
    OuhdDgq.jpg
     
  5. Apr 15, 2013 #4

    Mark44

    Staff: Mentor

    So y(t) = L-1(Y(s)) = L-1(1 /[(s - 1)2(s2 -8s + 20)]

    Break up the right side using partial fractions, and then you can take inverse Laplace transforms of them separately. This is how you should decompose the right side.

    $$ \frac{1}{(s - 1)^2(s^2 - 8s + 20)} = \frac{A}{s - 1} + \frac{B}{(s - 1)^2} + \frac{Cs + D}{s^2 - 8s + 20}$$

    When you figure out A, B, C, and D, check your work to make sure you haven't made an error. Then we can talk about the final step.
     
  6. Apr 15, 2013 #5
    So I was doing it right.. The answer was kinda big so I thought I made a mistake somewhere.

    KnJNvYS.jpg
     
  7. Apr 15, 2013 #6
    So is that right?
     
  8. Apr 15, 2013 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it is. Well done.
     
  9. Apr 16, 2013 #8

    Mark44

    Staff: Mentor

    In your post just before the one I'm quoting, you showed the solution you found. I assumed that you had checked your solution to verify that it works.
     
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