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Math Help - Doing something wrong

  1. #1
    Junior Member
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    Doing something wrong

    In connection with a thread posted earlier Anteeks was having problems with the equation:  y'-2y=3, y(0)=1.
    If you solve this with an integrating factor you get a factor of e^{-2x}.
    (e^{-2x}*y)'=3e^{-2x}
    e^{-2x}y=-\frac{3}{2}e^{-2x}+C
    y=-\frac{3}{2}+ce^{2x}
    If you plug in the initial condition, c=\frac{5}{2}, and this satisfies the original equation.
    However, Mr Fantastic pointed out that the equation was seperable, and indeed it is. But when I solve it that way:
    y'=2y+3
    \frac{y'}{2y+3}=^{dx}
    \ln({2y+3})=x+C
    2y+3=ce^x
    This does not satisfiy the original equation, so can't be right, but I don't see what I'm doing wrong.
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  2. #2
    A Plied Mathematician
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    One incorrect step was this:

    \displaystyle\int\frac{dy}{2y+3}=\ln|2y+3|, which is incorrect.

    You have to do a u=2y+3, du/2=dy substitution thus:

    \displaystyle\int\frac{dy}{2y+3}=\frac{1}{2}\int\f  rac{du}{u}=\frac{1}{2}\,\ln|u|=\frac{1}{2}\,\ln|2y  +3|.

    How does that change things?
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  3. #3
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    Thanks.
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  4. #4
    A Plied Mathematician
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    You're welcome. Does that enable you to finish the problem?
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