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D.E. Reduction of Order: Can't integrate!

  1. Feb 24, 2013 #1
    Last edited by a moderator: Feb 24, 2013
  2. jcsd
  3. Feb 24, 2013 #2

    rock.freak667

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    For your shortcut formula to work, what form must your DE be in i.e. how would you need to rewrite it as?
     
  4. Feb 24, 2013 #3
    oh yea. I need to divide everything by 6x^2. Thanks.
     
  5. Feb 24, 2013 #4

    rock.freak667

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    You need to not only divide by 6, but by x2 as well :wink:
     
  6. Feb 24, 2013 #5
    it still won't integrate..

    3Hnc0mU.jpg
     
  7. Feb 24, 2013 #6

    rock.freak667

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    By virtue of one solution being y = x^(1/2) you can say that x > 0.

    So that |x| = x

    If you remember how |x| is defined:

    |x| = -x for x<0
    |x| = x for x > 0
     
  8. Feb 24, 2013 #7
    So going on that concept, are these two answers correct?

    yZWlgYc.jpg
     
  9. Feb 24, 2013 #8

    rock.freak667

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    For 9: The actual integration is correct, however, your formula is wrong as you are multiplying by √2 instead of √x (y1).

    Number 10 should be correct. Not quite sure how you deduced the integral was equal to x2sin(lnx) but it is correct.
     
  10. Feb 24, 2013 #9
    Thanks. I fixed #9. For #10 I just used the Ti-nSpire CAS.
     
  11. Feb 24, 2013 #10

    rock.freak667

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    AH okay then. But for other integrals just make sure you can do them by hand if you ever need to.

    Also when you are putting together your general solution, you can combine the constants so if you have y=c1x+25c2x2, you can rewrite it as y=c1x+c3x2.
     
  12. Feb 24, 2013 #11
    noted. thanks
     
  13. Feb 25, 2013 #12
    What is the "largest interval of definition"? (0,inf)?
     
  14. Feb 25, 2013 #13

    rock.freak667

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    I believe that would be it as x > 0 would make the equation valid.
     
  15. Feb 25, 2013 #14
    What are the steps you go through to answer that part of the question? You look in the denominator of every step to try to see if any x value would make the equation undefined? or do you just see if any x value would make the equation undefined for the final answer? or do you see if any x value would make the equation undefined for y1 and y2?
     
  16. Feb 25, 2013 #15

    rock.freak667

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    Yes, you would check to see where the function exists. The function will not exist for x=0 as seen in the solution of y2 and not exist if x < 0 as seen in y1 and y2.
     
  17. Feb 25, 2013 #16
    ok. So to be specific, you only look at the y1 and y2 when checking where the function exists, not the any of the work done to get the solutions.
     
  18. Feb 25, 2013 #17

    rock.freak667

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    Basically.
     
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