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Help with Integrating Factor

  1. Feb 14, 2012 #1
    I am really struggling with proving a ODE by means of using the integrating factor method.
    My original problem was a Laplace transform
    q'+2q=5sin(t) where q(0)=0
    I believe i have got the correct naswer for this as being:- q= e^-2t +2sint-cost
    I just need to confirm this i have my integrating factor as e^2x but after that i am not really sure where to go next.
    I am surrounded by piles of paper with varying answers on, non of which are the same as the one above.
    Please please help someone.
     
  2. jcsd
  3. Feb 15, 2012 #2

    tiny-tim

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    welcome to pf!

    hi hurcw! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    yes, you multiply the orignal equation by e2t, giving:

    e2tq' +2qe2t = 5e2tsin(t)​

    that's the same as:

    (qe2t)' = 5e2tsin(t)​

    now you integrate both sides:

    qe2t = ∫ 5e2tsin(t) dt​

    and finally multiply the RHS (after the integration, and don't forget the constant!) by e-2t, to give you q :smile:
     
  4. Feb 15, 2012 #3
    Thanks for the reply if i integrate 5e^2t sin(t) do i not just get the same +C, and then if i multiply by e-2t it cancels out the e2ttso i am left with 5sin(t)+ C.
    Or am i being completely retarded which is a major possibility
     
  5. Feb 15, 2012 #4

    tiny-tim

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    nooo :redface:

    d/dt(5e2tsint) = 10e2tsint + 5e2tcost :wink:
     
  6. Feb 15, 2012 #5
    Ahhh i see.....i think,
    I then multiply this by e(-2t)
    Which cancels out the e(2t) am i right?
    i still dont see how i end up at my original answer?
     
  7. Feb 15, 2012 #6

    tiny-tim

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    show us what you've done :smile:
     
  8. Feb 15, 2012 #7
    Right here it is:-

    q'+2q=5sin(t)
    IF= e(2t) , Q=5sin(t)
    e(2t)q'+2e(2t)q=5e(2t)sin(t)
    d/dt2(e(2t)q)=5e(2t)sin(t)

    2(e(2t)q)=∫5e(2t)sin(t).dt

    = 10e(2t)sin(t)+5e(2t)cos(t)+C

    e(2t)q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2

    q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2e(2t)

    I then put in my original boundary condition of q(0)=0 to find C

    And end up with C=-5
    which kind of tallies up because q = 10e(2t)sin(t)+5e(2t)cos(t)+C

    0 = (0x1)+(5x1)-5
    Transposed 5 = 5
    Hows this look so far ??
     
    Last edited: Feb 15, 2012
  9. Feb 16, 2012 #8

    tiny-tim

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    (just got up :zzz: …)
    no, that last line is wrong …

    if you differentiate it, you don't get 5e(2t)sin(t) :redface:

    (also, that "2" on the far left came from nowhere)
     
  10. Feb 16, 2012 #9
    The '2' is from the original 2q.
    I don't get what you mean about my differentiation at the end, i have not differentiated yet.
    Where am i going wrong ?
    So is it this line that is wrong? q=(5(2e(2t)sin(t)+e(2t)cos(t))+C/2e(2t)
    Should it be q=5(2e(2t)sin(t)+e(2t)cos(t))+C/e(2t)
    which would cancel down to q=5(e(2t)sin(t)+cos(t))+C wouldn't it ?????
     
    Last edited: Feb 16, 2012
  11. Feb 18, 2012 #10
    Can anyone offer any more assistance with his please, all the help so far has been greatly appreciated.
     
  12. Feb 19, 2012 #11

    tiny-tim

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    hmm … let's rewrite that so that it's readable …
    no your ∫ was wrong (there should be a minus in the first line),

    and because you've put your brackets in the wrong place, there are errors in the next two lines also :redface:

    you must write these proofs out more carefully and in full (ie without taking short-cuts by missing out lines)​
     
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