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Solve the system using differential operators.

  1. May 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve the system using differential operators. Determine the # of arbitrary constants and then compare to your solution.

    2. Relevant equations
    D substitution: replace x' with Dx and y' with Dy

    3. The attempt at a solution
    x2L0r47.jpg

    I have the solution to this one, but I'm working another one like it and I don't know how I got the 4 + 8t - 2. Please help me understand what I did. We did this in class and I was half asleep. Final exam is tomorrow morning, so time is urgent. Thanks for your time. -David
     
  2. jcsd
  3. May 7, 2015 #2

    SammyS

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    What is meant by (D+2)(4t) ?
     
  4. May 7, 2015 #3
    (D+2) was multiplied to the whole equation on top and (2D+1) was multiplied by the one on the bottom so that I could get rid of x.
     
  5. May 7, 2015 #4

    SammyS

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    Yes, I know why you "multiplied" by (D+2) and and (2D+1) .

    To help you answer your question in post #1, my question to you was:
    What is meant by (D+2)(4t) ?​

    Simpler;
    What is D(4t) ?
    What is 2(4t) ?​
     
  6. May 8, 2015 #5
    I have no clue and I'm not a betting man, but if I was, I'd say that has something to do with the 8t. I have looked all over for a solution or a similar example. I'm probably just missing some small detail, but to be blunt, I'm dead tired, it's 12am, my exam is at 8, and I still have half a study guide to go over so I have given up as of now.
     
  7. May 8, 2015 #6

    SammyS

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    What does D stand for ?
     
  8. May 8, 2015 #7
    Derivative

    Edit: ? I question myself these days.
     
  9. May 8, 2015 #8

    SammyS

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    yes. derivative.

    And in this case, it's derivative with respect to t.

    Now again:
    What is meant by (D+2)(4t) ?​
     
  10. May 8, 2015 #9
    So you take the derivative of 4t to be equal to 4, and your (2)(4t) to be 8t?

    Do you also get the -2 from the (D+2)? just a thought
     
  11. May 8, 2015 #10

    SammyS

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    No. How could that be?

    The other equation has (2D + 1)(2) on the right side .

    What does that give?
     
  12. May 8, 2015 #11
    if you multiply both out you just get 4Dt+8t and 4D + 2 I'm not sure where you're going with this.
     
  13. May 8, 2015 #12

    Mark44

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    I'm not surprised that you don't know how you got 4 + 8t -2, since the work just before that is very disorganized. The line just above that is not an equation, and it's not clear wat that line means, or how it's related to the lines above it.

    The main reason we ask people NOT to merely post a snapshot of their work is that we can't insert comments right at the point where things go wrong, but have to describe the point in the image. Although I follow what you're doing in part, it's difficult for me to explain what you're doing (and why) without counting your lines and describing them by line number. In short, posting a photo makes it harder for us to help you. I realize that you're a new member, but please take some time to familiarize yourself with either the BBCode markup (available in part from the X2 symbol in the green strip across the top of the input pane) or using LaTeX. The INFO menu at the top of the screen has a link to a page of Help/How-To topics, including BBCode markup and and a LaTeX tutorial.

    In lines 5 and 6 you are not multiplying by D + 2 and 2D + 1, respectively -- you are applying the D + 2 operator to both sides of the first equation and the 2D + 1 operator to both sides of the second equation. The right side of the first equation is (D + 2)(4t). The right side of the second equation is (2D + 1)(2). The first expression here should not be thought of as "(D + 2) times 4t", but, rather, as "(D + 2) of 4t". IOW, you are applying the operator take-the-derivative-with-respect-to-t-of-plus-2-times 4t. And similar for the second equation, but with the operator 2D + 1.

    Next, you are subtracting the equation in line 6 from the equation in line 5, getting rid of all the terms that involve y. What are you left with at that point?
     
  14. May 8, 2015 #13
    AHHHHHH This makes so much more sense now. Thank you very much, both of you. I understand the Sammy's approach, but considering I had no idea what was going on it was hard to figure it out on my own. My teacher is Chinese and has a hard time relating things to us. I looked at this more algebraically and had not considered the derivatives at all. Thanks guys. If I can get partial credit, maybe it'll save me. It gives you the 4 + 8t, then you subtract the 2 from the second equation. I'll try to refrain from posting snapshots in the future. Thanks again

    My spirit has been released
    DoveRelease.48122820_std.jpg
     
    Last edited: May 8, 2015
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