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Math Help - using a transformation to simplify a differential equation

  1. #1
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    using a transformation to simplify a differential equation

    Hi all,
    Is it just me or is there something strange going on here.

    Consider the following differential equation:

    0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]

    Consider the transformation (x<1/2)

     h(x)=\frac{1/2-f(x)}{1/2-x}.

    Then the differential equation is supposed to become:

    h'(2h-x)(\frac{1}{2}-x)=2h(h-2)



    However, when I did it I got:

    h'(2h-3)(\frac{1}{2}-x)=2h(h-2).


    I did the following:

    Calculate f'(x):

    f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)
    => f'(x)=h(x)-h'(x)(\frac{1}{2}-x)

    Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

    0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]
    0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]
    0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]
    h'(\frac{1}{2}-x)(3-2h)=2h(2-h)
    h'(\frac{1}{2}-x)(2h-3)=2h(h-2)


    So what am I missing here (I'm 99% sure it's me who's wrong)?

    Thanks very much!
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  2. #2
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    Quote Originally Posted by honda View Post
    Hi all,
    Is it just me or is there something strange going on here.

    Consider the following differential equation:

    0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]

    Consider the transformation (x<1/2)

     h(x)=\frac{1/2-f(x)}{1/2-x}.

    Then the differential equation is supposed to become:

    h'(2h-x)(\frac{1}{2}-x)=2h(h-2)



    However, when I did it I got:

    h'(2h-3)(\frac{1}{2}-x)=2h(h-2).


    I did the following:

    Calculate f'(x):

    f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)
    => f'(x)=h(x)-h'(x)(\frac{1}{2}-x)

    Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

    0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]
    0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]
    0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]
    h'(\frac{1}{2}-x)(3-2h)=2h(2-h)
    h'(\frac{1}{2}-x)(2h-3)=2h(h-2)


    So what am I missing here (I'm 99% sure it's me who's wrong)?

    Thanks very much!
    Dear honda,

    I did the same substitution and ended up with your answer. I think your answer is correct.
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