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Solve y''+9y=delta(t-pi)

  1. Jul 5, 2008 #1
    Solve the differential equation

    y''+9y=delta(t-pi)

    that fulfills the initial condition y(0)=y(0)=1. Answer by giving the value for y((14*pi)/9). The answer can be given by a fraction a/b.

    3. The attempt at a solution

    I will submit my attempt for a solution as an attachment shortly.
     
  2. jcsd
  3. Jul 5, 2008 #2
    I am now at the stage where I need to get the inverse transforms but I don't remember how...
     

    Attached Files:

  4. Jul 6, 2008 #3
    I can't figure out how to get the inverse transforms...any ideas?
     
  5. Jul 6, 2008 #4

    Defennder

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    There is a mistake on the 3rd line. [tex]sY(0)[/tex] is not 1. It's just s. Also [tex]s^2 + 9 [/tex] is not [tex](s+3)^2[/tex].
     
  6. Jul 6, 2008 #5
    Defennder thanks for pointing that out. So now I have the following:
     

    Attached Files:

  7. Jul 7, 2008 #6
    I just don't understand how these three fractions can give me y at the end. Is the approach of inverse transform incorrect? Cause, just by looking at s/(s^2+9) I can transform that to cos(3t) but what do I do with the rest. How can these three "add up" to give me the fraction a/b?
     
    Last edited: Jul 7, 2008
  8. Jul 7, 2008 #7

    Defennder

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    Your attachment hasn't been approved yet. Try hosting it on imageshack.us and linking the picture. That way you won't have to wait for approval from the mentors.
     
  9. Jul 7, 2008 #8
    Thanks for letting me know...I hope this works:)
    http://img57.imageshack.us/my.php?image=mathprobkv7.jpg
     
  10. Jul 7, 2008 #9

    Defennder

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    So far so good. I got the same answer. To inverse-Laplace transform the RHS into a function of t again, you'll need the laplace transform of:

    [tex]L[\cos (\omega t)][/tex]
    [tex]L[\sin (\omega t)][/tex]
    [tex]L[f(t-a)u(t-a)][/tex] where u(t-a) is the unit step function or Heaviside function. Look this up and inverse-transform them individually.
     
  11. Jul 7, 2008 #10
    The Laplace Transform of cos(at) is s/(s^2+a^2)
    The Laplace Transform of sin(at) is a/(s^2+a^2)
    and Laplace Transform of f(t-a)u(t-a) is exp(-Ts)F(s)

    Now I will try to inverse-transform them individually.
     
  12. Jul 7, 2008 #11

    Defennder

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    All ok except this one. Probably a typo.
     
  13. Jul 7, 2008 #12
    Oh yeah and Laplace Transform of f(t-a)u(t-a) is exp(-a*s)F(s)
    is that correct now?
     
  14. Jul 7, 2008 #13

    Defennder

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    Yeah, it is. You should be able to solve for y now.
     
  15. Jul 7, 2008 #14
    I have the following...

    http://img71.imageshack.us/img71/4016/mathprob1na5.jpg

    But I'm not sure about that Heaviside function term. I really don't know what to do with (s^2+9) in denominator? Isn't there a jump (discontinuity in Heaviside functions)? I have not y yet. If I add these three is that how I get y? What do I have to do about that F(s) that popped up? And do I just plug in t=((14*pi)/9) at the end when I get y? I am thankful for any suggestions...
     
    Last edited: Jul 7, 2008
  16. Jul 7, 2008 #15
    After some thinking about the Heaviside term I came to the conclusion that the last term in the attachment above should have an inverse Laplace transform of: u(t)f(t-pi)*(1/3)*sin(3t)

    If this is correct (please object if it is not) as I add all three terms I should get:

    y(t)=u(t)f(t-pi)*(1/3)*sin(3t)+(1/3)*sin(3t)+cos(3t)

    According to my understanding I should now plug in t=((14*pi)/9) into y(t) as shown above:

    y(t=((14*pi)/9))=u((14*pi)/9))f((14*pi)/9)-pi)*(1/3)*sin(3*(14*pi)/9))+(1/3)*sin(3*(14*pi)/9))+cos(3*(14*pi)/9))
     
  17. Jul 7, 2008 #16

    Defennder

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    This isn't right. Remember that both the Heaviside function and the function f(t) is shifted by -a. So u(t) is not it. f(t) here, which I take to be 1/3sin(3t) is also not shifted correctly.
     
  18. Jul 7, 2008 #17
    I see...is this what you mean?

    u(t-a)f(t-a)*(1/3)*sin(3t-a) Is the a-value for the shift pi?

    If this is true, will this be the expression for y:

    y=u(t-pi)f(t-pi)1/3sin(3t-pi)+1/3sin(3t)+cos(3t)

    How will I evaluate the first term in the expression above numerically?
     
    Last edited: Jul 7, 2008
  19. Jul 7, 2008 #18
    I plugged in t=((14*pi)/9) into cos(3t) & 1/3sin(3t)...

    cos(3t)=-0.5
    1/3sin(3t)=-0.32827

    I still don't know what to to about the heaviside-term. How can I evaluate that numerically? And how do I get a fraction a/b from all this? Any ideas or suggestions?
     
  20. Jul 7, 2008 #19
    Can't anyone shed more light on this problem?
     
  21. Jul 7, 2008 #20

    Defennder

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    That's not the inverse transform of [itex]\frac{e^{-\pi s}}{s^2+9}[/itex]. Remember that f(t-a) is not some unknown function; it's just the inverse Laplace transform of [tex]\frac{1}{s^2+9}[/tex], which is shifted by -a. What this means is that in place of t in the original inverse transform of [tex]\frac{1}{s^2+9}[/tex] you now must write [itex]t-a[/itex]. And after that is done, write u(t-a) next as a term multiplied to it.

    As for the meaning of the Heaviside function u(t-a), note that all it does is ensure that the graph of f(t)u(t-a) would be 0 for t<a. For t>a, the graph would simply be f(t-a). You can treat u(t-a)f(t) as a shorthand for writing:

    [tex]f(t)u(t-a) = \begin{array}{c}0 \ \mbox{for} \ t<a\\f(t) \ \mbox{for} \ t>a \end{array}[/tex]
     
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