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Math Help - A+b=a'/b'

  1. #1
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    A+b=a'/b'

    Find two real functions whose sum is the ratio of their first derivatives, i.e. solve:

    A(x)+B(x)=\frac{A'(x)}{B'(x)}

    Or prove no solutions exist.

    (This is a problem that cropped up in another thread: http://www.mathhelpforum.com/math-he...tml#post345253)
    Last edited by Media_Man; August 2nd 2009 at 06:13 AM. Reason: referencing other thread
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  2. #2
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    There are an infinite number of solutions aren't there ?

    Just solve as a first order ode in  A(x).

    Rewrite as

     A'(x) - B'(x)A(x) = B(x)B'(x) ,

    multiply by the integrating factor

     e^{-B(x) }

    and the equation can be written as

     (A(x)e^{-B(x)})' = B(x)B'(x)e^{-B(x)}

    so

     A(x)e^{-B(x)} = \int B(x)B'(x)e^{-B(x)} dx

    and

     A(x) = e^{B(x)}\int B(x)B'(x)e^{-B(x)} dx .

    Now just choose something convenient for  B(x) .

    An easy choice would be  B(x) = x after which an integration by parts would lead to  A(x) = Ce^{x} - x - 1 .
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  3. #3
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    Quote Originally Posted by BobP View Post
    There are an infinite number of solutions aren't there ?

    Just solve as a first order ode in  A(x).

    Rewrite as

     A'(x) - B'(x)A(x) = B(x)B'(x) ,

    multiply by the integrating factor

     e^{-B(x) }

    and the equation can be written as

     (A(x)e^{-B(x)})' = B(x)B'(x)e^{-B(x)}

    so

     A(x)e^{-B(x)} = \int B(x)B'(x)e^{-B(x)} dx

    and

     A(x) = e^{B(x)}\int B(x)B'(x)e^{-B(x)} dx .

    Now just choose something convenient for  B(x) .

    An easy choice would be  B(x) = x after which an integration by parts would lead to  A(x) = Ce^{x} - x - 1 .
    We can actually go further and integrate fully giving

     <br />
A = c_1 e^{B} - B - 1<br />
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