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Math Help - dy is to dx as y is to x

  1. #1
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    dy is to dx as y is to x

    I'm looking for functions that satisfy:

    \frac{dy}{dx}=\frac{y}{x}k

    (k being a constant)

    So far I am aware that y=ax^k works, but are there any other functional forms out there that would also work?

    Thanks
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  2. #2
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    Quote Originally Posted by rainer View Post
    I'm looking for functions that satisfy:

    \frac{dy}{dx}=\frac{y}{x}k

    (k being a constant)

    So far I am aware that y=ax^k works, but are there any other functional forms out there that would also work?

    Thanks
    This is a separable equation, so you would integrate (and don't forget the constant of integration).
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    Quote Originally Posted by wonderboy1953 View Post
    This is a separable equation, so you would integrate (and don't forget the constant of integration).
    So, integrating \frac{dy}{dx}=\frac{y}{x} k I get y=ky\ln{x}+c

    Solving for y I get: y=\frac{c}{1-k\ln{x}}

    This yields the following relationship:

    \frac{dy}{dx}=\frac{y^2}{x}\frac{k}{c}

    Which is tantalizingly close to the proportion I was looking for, but not quite.

    I don't mind the "c" being in there, but is there some way to get the y to not be squared?
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    I get |y| = e^{k\ln|x| + C} = e^{\ln|x^k|} \cdot e^C = Px^k \implies y = \pm P|x|^k where P and k are constants for the integral which is what you got in the OP
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  5. #5
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    Quote Originally Posted by rainer View Post
    So, integrating \frac{dy}{dx}=\frac{y}{x} k I get y=ky\ln{x}+c

    Solving for y I get: y=\frac{c}{1-k\ln{x}}
    NO, no, no! That's not what "separating" means. You cannot treat the "y" on the right as if it were a constant while integrating dy on the left.
    Instead you need to actually separate x and y.
    \frac{dy}{y}= \frac{k}{x} dx

    Now, integrate both sides to get ln(|y|)= k ln(|x|)+ C= ln(|x|^k)+ C and take the exponential of both sides: y= |x|^ke^C= c|x|^k where c= e^C.

    This yields the following relationship:

    \frac{dy}{dx}=\frac{y^2}{x}\frac{k}{c}

    Which is tantalizingly close to the proportion I was looking for, but not quite.

    I don't mind the "c" being in there, but is there some way to get the y to not be squared?
    Yes, integrate correctly!
    Last edited by HallsofIvy; January 13th 2011 at 11:49 AM.
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    Quote Originally Posted by HallsofIvy View Post
    NO, no, no! That's not what "separating" means. You cannot treat the "y" on the right as if it were a constant while integrating dy on the left...

    ...integrate correctly!
    Oops!
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  7. #7
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    Good response! By the way, I note that I took the logarithm incorrectly on the right and have edited my previous response.
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    Hey guys, just one last question:

    When taking the integral k\int \frac{1}{x}\: dx why doesn't the k get factored out?

    I.e., to my naive mind it should be:

    k\int \frac{1}{x}\: dx=k(\ln{x}+C)

    Where you guys have:

    k\int \frac{1}{x}\: dx=k\ln{x}+C

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  9. #9
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    Notice that \displaystyle k(\ln{x} + C) = k\ln{x} + kC = k\ln{x} + C' where \displaystyle C' is some other constant.

    So it really doesn't matter how you write it.

    BTW I expect you need modulus signs inside the logarithm...
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  10. #10
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    oh yeah. Duh!
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