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Math Help - Integrating Factor Problem

  1. #1
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    Integrating Factor Problem

    I was wondering if anyone can help me in finding a general solution to this:

    <br />
x\prime(t) + 2tx(t) = t<br />

    I've tried using the integrating factor method but I must be getting it wrong or else it's the wrong method to use because the answer is weird...
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  2. #2
    Super Member Showcase_22's Avatar
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    I did it a different way:



    I haven't seen the notation you used before but I think i've written it down correctly. I also didn't use the integrating factor method. It just seemed easier to do it this way.
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  3. #3
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    Great, thanks!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by QuestionMark View Post
    I was wondering if anyone can help me in finding a general solution to this:

    <br />
x\prime(t) + 2tx(t) = t<br />

    I've tried using the integrating factor method but I must be getting it wrong or else it's the wrong method to use because the answer is weird...
    One way that was suggested was separation of variables.

    This is how you would do it with the integrating factor:

    x'(t)+2tx(t)=t

    Let P(t)=2t

    Thus, the integrating factor \varrho=e^{\int P(t)\,dt}=e^{2\int t\,dt}=e^{t^2}

    Now, when you multiply the equation through by \varrho, the left side always becomes the derivative of the product between the dependent variable (in this case, x), and the integrating factor.

    So we see that the equation now becomes \frac{d}{dt}\left[x\cdot e^{t^2}\right]=te^{t^2}

    Integrate both sides and you get x\cdot e^{t^2}=\tfrac{1}{2}e^{t^2}+C

    Now multiply both sides by e^{-t^2} to get the solution

    \color{red}\boxed{x(t)=\tfrac{1}{2}+Ce^{-t^2}}

    My solution and Showcase_22's are almost the same. However, he made a tiny mistake:

    \int\frac{\,dx}{1-2x}=\int t\,dt

    When you integrate, you get -\tfrac{1}{2}\ln|1-2x|={\color{red}\tfrac{1}{2}}t^2+C

    His solution in the end would then be x(t)=\tfrac{1}{2}\left[1-Ae^{-t^2}\right]

    Note that -\tfrac{1}{2}\cdot A=C, another constant.

    So we would then get the solution \color{red}\boxed{x(t)=\tfrac{1}{2}+Ce^{-t^2}}

    They now match.

    I hope this makes sense!

    --Chris
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  5. #5
    Super Member Showcase_22's Avatar
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    oh darn it!

    I haven't been bringing my A-game lately, I should stop doing maths and msning at the same time! lol

    anyway, thanks for pointing out where I went wrong. I can't believe I did that! =O
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