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Math Help - First order DE

  1. #1
    MHF Contributor arbolis's Avatar
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    First order DE

    I solved the DE x'=\frac{1}{t}. I got x=\ln (t)+C.
    I've been asked to determine C if t=-1.
    So I think that the solution of the DE should be x=\ ln |t|+C so that I can reach C=2.
    I just want to be sure we can talk about negative values of t.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    I solved the DE x'=\frac{1}{t}. I got x=\ln (t)+C.
    I've been asked to determine C if t=-1.
    So I think that the solution of the DE should be x=\ ln |t|+C so that I can reach C=2.
    I just want to be sure we can talk about negative values of t.
    Yes, you can do that because \int\frac{\,dx}{x}=\ln{\color{red}|}x{\color{red}|  }+C. With that, you're able to find the constant C.
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  3. #3
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    Negative?

    Your t can be negative, because the anti derivative of 1/t is ln|t| + C. The variable is an absolute value, which of course means that if it is negative you just make it positive

    I am not entirely sure how to solve this problem because you did not give an x component to go with the t. Usually you are given x, or are can find it out. The ln|-1| is 0, so that would mean C would equal whatever x value you find. Hope that explains it
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Orbent View Post
    Your t can be negative, because the anti derivative of 1/t is ln|t| + C. The variable is an absolute value, which of course means that if it is negative you just make it positive

    I am not entirely sure how to solve this problem because you did not give an x component to go with the t. Usually you are given x, or are can find it out. The ln|-1| is 0, so that would mean C would equal whatever x value you find. Hope that explains it
    Yes sorry! I forgot to say that I was given x(-1)=2. Hence my value of C. Now I know why the absolute value is extremely important, I won't forget it ever.
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