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Math Help - Intergrating Factors

  1. #16
    Newbie
    Joined
    Feb 2009
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    I take it now all i need to do is
    50\ \theta\ e^{t/10}\ =\ 50e^{t/10} - 25te^{t/10} - (-250e^{t/10}) - 240
     \theta\ = (\ 50e^{t/10} - 25te^{t/10} - (-250e^{t/10}) - 240) / (50 e^{t/10})
    So then i sub t=2 in to give me theta when t=2 seconds?

     \theta\ = (\ 50e^{2/10} - (25*2)e^{2/10} - (-250e^{2/10}) - 240) / (50 e^{2/10})
    is that correct?
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  2. #17
    MHF Contributor
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    Quote Originally Posted by jamm89 View Post
    is that correct?
    You have at least two ways of checking. Differentiate the 'whole equation' to see if you can arrive at the original given equation. Also, is the answer realistic? Helps if you know was it heating or cooling? Maybe you don't know.
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