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Help me solve this differential equation

  1. Apr 21, 2015 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    find the inverse laplace transform.

    1/(s^3 + 7s)

    2. Relevant equations
    sin(kt) = k/(s^2 + k^2)
    cos(kt) = s/(s^2 + k^2)
    3. The attempt at a solution

    so this one is different from all the others I have done because it involves an imaginary number and I am not sure what the rule is that changes the problem.
    somehow the answer becomes 1/7 - cos(sqrt(7)*t ) / 7

    but I am getting positive sine instead of negative cosine and 7i in the denominator instead of 7. does the i reverse sine to cosine or something?
     
  2. jcsd
  3. Apr 21, 2015 #2

    Simon Bridge

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    ... try taking the Laplace transform of the answer and see if it matches the question.
     
  4. Apr 21, 2015 #3

    grandpa2390

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    Well... considering it is the right answer... it should match the question. What is the method for getting that answer.
     
  5. Apr 21, 2015 #4

    Simon Bridge

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    It can be instructive to check that the "correct answer" matches the question ... did you try? Did you try the transform on your answer?

    Going forward:
    Did you try completing the square in s^2 in the denominator.
     
  6. Apr 21, 2015 #5

    grandpa2390

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    I don't know what that means.

    I set up my partial fraction. A(s^2+7) + B(s) = 1
    s=0 and sqrt(-7)
    A= 1/7 and B= 1/sqrt(-7)

    1/(7s) + 1/(sqrt(-7)*(s^2+7))

    1/7 * 1/s = 1/7
    1/sqrt(-7)sqrt(7) * sqrt(7)/(s^2 + 7) = 1/(7i) * sin(sqrt(7)t)

    1/7 + sin(sqrt(7)t)/7i
    but that is wrong I should get 1/7 - cos(sqrt(7)t)/7
     
    Last edited: Apr 21, 2015
  7. Apr 21, 2015 #6

    Simon Bridge

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    Try getting the denominator in form: ##(s^2 + a^2)^2##

    Put it another way: please show your working.
     
  8. Apr 21, 2015 #7

    grandpa2390

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    there I posted my work.
     
  9. Apr 21, 2015 #8

    grandpa2390

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    so you are saying it is ok to multiply the initial problem by s/s and then I won't have to worry about i.

    edit: no that didn't work

    I do not have a clue.
     
    Last edited: Apr 21, 2015
  10. Apr 21, 2015 #9

    Simon Bridge

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    ... In the partial fractions decomposition: don't you need a Bs+C in the numerator of the quadratic denominator?
    I found it helped to put a^2=7 and just write everything out in terms of a.
     
  11. Apr 21, 2015 #10

    grandpa2390

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    I'll try yahoo answers.
    that might be it then. I should have done Bs+C. and that would put the s in the numerator required for the cosine function. but will it get rid of the i?
     
  12. Apr 21, 2015 #11

    Simon Bridge

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    If I read you correctly: the "i" comes from the imaginary root of the quadratic - nothing to do with the decomposition.
    In general, the polynomial in the numerator should be, at most, one order less than the polynomial in the denominator.
     
  13. Apr 21, 2015 #12

    Ray Vickson

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    Use either (1) If ##f(t) \leftrightarrow g(s)## then ##\int_0^t f(\tau) \, d\tau \leftrightarrow \frac{g(s)}{s}##; or (2) a partial-fraction expansion of ##\frac{1}{s^3 + 7s}##.
     
  14. Apr 23, 2015 #13

    Simon Bridge

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    I had a think further - you tried a "partial fractions" of form:
    $$\frac{A}{s}+\frac{B}{s^2+7}$$... and discovered the relations for A and B were: $$As^2 + 7A +Bs = 1\\ \implies As^2=0,\; 7A=1,\; Bs=0$$ ... the first tells you that A=0 while the second tells you that A=1/7 ... which is an inconsistency, which should have been a hint that you'd done something wrong.
    By working out the roots of s^2+7 you were trying to find the values of s that would make the relations true(?) ... which is an OK strategy except that the decomposition is supposed to work for all values of s not just a few of them, so the strategy failed. How you incorporated the roots into the partial fractions is still beyond me - but I suspect it is enough to notice that you had not understood what the partial fractions method is supposed to do.
     
    Last edited: Apr 24, 2015
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