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Multi-Step linear systems problem - Build ODE, impulse response, systems balance

  1. Oct 21, 2009 #1
    I messed up on the title of the thread.. I had a few questions but ended up solving them on my own. The only question I have is how to solve this ODE. Thanks.




    1. The problem statement, all variables and given/known data

    Solve the ODE

    (dx/dt)= f(t) - .1x

    Where f(t) = 2+sin(t)


    2. The attempt at a solution

    I faintly remember doing these types of ODE's in diff eq class, but there was always a x/t term so it ended up being a nice equation. Not really sure how to approach one of these though.
     
  2. jcsd
  3. Oct 21, 2009 #2
    I learned this recently in my class. Try moving all the x factors to the left and see if you can do something there.
     
  4. Oct 21, 2009 #3
    I think its coming back to me.. I need to find an integrating factor

    I found that the integrating factor would be e-.1t


    If I multiply both sides by it and integrate I end up with:

    e-.1tx = -20e-.1t+int(sin(t)e-.1tdt)


    So I end up having the int(sin(t)e-.1tdt)

    Which isn't being too cool with letting me solve it.. any suggestions?
     
  5. Oct 21, 2009 #4
    You're on the right track.
    Remember you have a dx/dt on the left so the equation would be
    (dx/dt)+.1x= 2+sin(t)

    by turning this into a different notation you get:
    x'+.1x=2+sin(t)

    So then you can now multiply by the integrating factor and get
    e^(.1t)x'+.1e^(.1t)x=2+sin(t)

    Does the left side look familiar?
     
  6. Oct 21, 2009 #5

    Dick

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    Homework Helper

    Yes, I think it's coming back to you. I would say the integrating factor is for x'+x/10 is exp(t/10) though. Keep working on it. I think it will come back. To integrate sin(t)*exp(t/10) you integrate by parts. Twice. Is the memory coming back?
     
  7. Oct 21, 2009 #6
    As of right now I'm still stumped.. but I did realize I made a mistake with a sign.

    After integrating twice I end with:


    et/10x = 20et/10 + 10et/10sin(t) - 100et/10cos(t) - 100 [tex]\int[/tex]et/10sin(t)dt


    I think I might see something..

    [tex]\int[/tex]et/10sin(t)dt = et/10x-20et/10?


    Still a noob with the symbols :P


    Thanks for all your help so far!
     
  8. Oct 21, 2009 #7
    Try my way...
    e^(.1t)x'+.1e^(.1t)x=2+sin(t)

    The left side is just the product rule. So if you do this:
    d/dt(e^(.1t)*x)=2+sin(t)

    Then you can do a really simple integration. But you could also do it Dick's way.
    :)
     
  9. Oct 21, 2009 #8
    I'm all for the easy way (dont shoot me!) but what I don't see in your way is why you only multiplied one side by the integrating factor. Should you multiply the other side as well?
     
  10. Oct 21, 2009 #9

    Dick

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    No, what's an x doing in there? You have to integrate exp(t/10)*sin(t)dt by parts using d(-cos(t))=(sin(t))dt. Then you integrate by parts AGAIN and you get the integral of exp(t/10)*sin(t) back again. But with a different sign. Solve for the integral of exp(t/10)*sin(t). I think you've probably seen this before.
     
  11. Oct 21, 2009 #10
    Yeah sorry about that, I forgot. :eek:
    Multiply both sides by exp(.1t).

    The integration is harder now but its still the basic concept.
     
  12. Oct 21, 2009 #11
    So I redid my integration and got..

    et/10x = 20et/10 - et/10cos(t) - (1/10)et/10sin(t) - (1/100) [tex]\int[/tex]et/10sin(t)dt


    cmon brain dont fail me now.. i still dont see it arg
     
  13. Oct 21, 2009 #12
    Remember Integration by Parts has "tricks"?
    Try setting your equation to

    [tex]\int[/tex] exp(t/10)sin(t)dt =20et/10 - et/10cos(t) - (1/10)et/10sin(t) - (1/100)[tex]\int[/tex] et/10sin(t)dt
     
  14. Oct 22, 2009 #13
    (I made a mistake with the 1/10 term, should be +)

    anyways


    OK!! I think I figured it out..


    You can sub the integral parts for a variable.. in this case i'll call it Y

    So I solve for Y, plug it back in where the integral was.

    for y I ended up with

    Y = (100/101)(20et/10 - et/10cos(t) + (1/10)et/10sin(t))

    So then u plug it back into where the integral was, exp functions end up canceling out and u get your answer for x

    right?



    ah, what a complicated answer for such a simple looking problem. math is the master of disguise :)
     
    Last edited: Oct 22, 2009
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