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Math Help - Solve (x-1)dx-(3x-2y-5)dy= 0.

  1. #1
    Member Miss's Avatar
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    Solve (x-1)dx-(3x-2y-5)dy= 0.

    hi

    i want to solve

    (x-1)dx-(3x-2y-5)dy=0

    But I could not classify it
    any help?
    Last edited by mr fantastic; March 23rd 2011 at 04:40 AM. Reason: Re-titled.
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  2. #2
    A Plied Mathematician
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    When you have linear functions multiplying the differentials like that, what you want to do is have a substitution based on the point of intersection of the two lines (if they are not parallel, as here). So, the left-hand line is x - 1 = 0, and the right-hand line is 3x - 2y - 5 = 0. Obviously, x = 1, and you can find the corresponding y. So let's call the point of intersection (x_{0},y_{0}). You want to do the substitution t=x-x_{0}, and z=y-y_{0}. The resulting DE will be of a form that you know, I think. What do you get?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Miss View Post
    hi

    i want to solve

    (x-1)dx-(3x-2y-5)dy=0
    @ Ackbeet. Interesting. I've never head of that before, but it looks like a keeper.

    I had another idea. Let's turn the usual method on its head: Solve this as x(y) instead of y(x):

    I get \displaystyle (x - 1)\frac{dx}{dy} - 3x = -2y - 5
    which is a much more recognizable form than the original.

    -Dan

    Uh. D*mn. I guess this one doesn't work out that well.
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  4. #4
    Member Miss's Avatar
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    @ Ackbeet:

    Actually, the substitutions should be t=x+x_0 & z=y+y_0.
    I know that kind of ODEs, but I forgot it.


    @ topsquark:

    Well, what can be done for the new form?
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  5. #5
    A Plied Mathematician
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    Quote Originally Posted by Miss View Post
    @ Ackbeet:

    Actually, the substitutions should be t=x+x_0 & z=y+y_0.
    Well, you can certainly try that substitution, but I don't think it'll do for you what my substitution will do for you. Good luck!

    I know that kind of ODEs, but I forgot it.


    @ topsquark:

    Well, what can be done for the new form?
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  6. #6
    A Plied Mathematician
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    Quote Originally Posted by topsquark View Post
    @ Ackbeet. Interesting. I've never head of that before, but it looks like a keeper.
    Yeah, I once saw someone on MHF mention that method, and then I saw it in Tenenbaum and Pollard, and so I practiced it quite a few times. A good reason to supplement Zill with T&P!
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Miss View Post

    @ topsquark:

    Well, what can be done for the new form?
    Apparently not much. It looks all nice, but even WolframAlpha had to think about it for a minute. (I put the link in the addendum to my first post) It looks similar to a problem I once saw, but apparently I'm remembering wrong.

    -Dan
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  8. #8
    Member Miss's Avatar
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    I found the method is explained here by the member General :

    http://www.mathhelpforum.com/math-he...on-163550.html
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  9. #9
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    Quote Originally Posted by Miss View Post
    I found the method is explained here by the member General :

    http://www.mathhelpforum.com/math-he...on-163550.html
    Right; thanks for the link. If you examine his substitution closely, you'll see that it's the same as mine. He just has the x_{0},y_{0} on the other side of the equation. Equivalently, my substitution is t+x_{0}=x, and z+y_{0}=y.
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