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Math Help - Elimination of Arbitrary Constants

  1. #1
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    Elimination of Arbitrary Constants

    How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

    The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

    Here is my attempt:

    y = ax^2 -1

    dy/dx = 2ax

    (dy/dx) (1/2x) = a

    therefore y = (dy/dx) (1/2x)x^2 - 1

    Where did I go wrong?
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  2. #2
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    Quote Originally Posted by sparky View Post
    How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

    The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

    Here is my attempt:

    y = ax^2 -1

    dy/dx = 2ax

    (dy/dx) (1/2x) = a

    therefore y = (dy/dx) (1/2x)x^2 - 1

    Where did I go wrong?
    Stopping too soon ?

    \displaystyle\ y=\frac{dy}{dx}\left(\frac{1}{2x}\right)x^2-1

    2y=2\displaystyle\left[\frac{dy}{dx}\left(\frac{x}{2}\right)-1\right]
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by sparky View Post
    How do I eliminate the arbitrary constant in the equation y = ax^2 -1?

    The answer at the back of the book is 2y = x(dy/dx)-2 however I don't understand how they arrived at this.

    Here is my attempt:

    y = ax^2 -1

    dy/dx = 2ax

    (dy/dx) (1/2x) = a

    therefore y = (dy/dx) (1/2x)x^2 - 1

    Where did I go wrong?
    You didn't do anything wrong you just need to keep going. Your answer is equivalent.

    \displaystyle y=\frac{dy}{dx}\left( \frac{1}{2x}\right)x^2-1 \iff y=\left( \frac{x}{2} \right) \frac{dy}{dx}-1 \iff 2y=x\frac{dy}{dx}-2

    Edit: too slow
    Last edited by TheEmptySet; January 31st 2011 at 11:24 AM. Reason: too slow
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  4. #4
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    Thanks for your replies TheEmptySet and Archie Meade.

    Your workings have brought me closer to understanding this.

    Why do we have to multiply both sides by 2?
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  5. #5
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    Only to remove the "inconvenient" fraction (a half) on the right, nothing more!
    Even without going that far, you were still correct where you finished up.
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  6. #6
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    Ok great, thanks a lot Archie Meade and TheEmptySet
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