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Math Help - Solving ODE with IVP

  1. #1
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    Solving ODE with IVP

    Ok, I'm stuck! I've tried everything and can't seem to solve this problem!

    dy/dt+2y=1 y(0)=5/2; and find the value of the function t=3seconds

    I first tried: dy=1-2y dt
    then dy/1-2y=dt
    next integrate both sides: u=1-2y u'=-2 dx=-1/2du=t
    so after integrating: -1/2ln(1-2y)=t+C
    When I plug in the IVP, I'm getting a negative log, So now I'm stuck!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by latavee View Post
    Ok, I'm stuck! I've tried everything and can't seem to solve this problem!

    dy/dt+2y=1 y(0)=5/2; and find the value of the function t=3seconds

    I first tried: dy=1-2y dt
    then dy/1-2y=dt
    next integrate both sides: u=1-2y u'=-2 dx=-1/2du=t
    so after integrating: -1/2ln(1-2y)=t+C
    When I plug in the IVP, I'm getting a negative log, So now I'm stuck!
    \int\frac{\,du}{u}=\ln{\color{red}\left|\right.}u{  \color{red}\left.\right|}+C.
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  3. #3
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    so the -1/2 ln(u) +C isn't correct? Thanks! I found your tutorials very helpful!
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  4. #4
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    Quote Originally Posted by latavee View Post
    Ok, I'm stuck! I've tried everything and can't seem to solve this problem!

    dy/dt+2y=1 y(0)=5/2; and find the value of the function t=3seconds

    I first tried: dy=1-2y dt
    then dy/1-2y=dt
    next integrate both sides: u=1-2y u'=-2 dx=-1/2du=t
    so after integrating: -1/2ln(1-2y)=t+C
    When I plug in the IVP, I'm getting a negative log, So now I'm stuck!
    dt = \frac{dy}{1 - 2y}.

    There are two approaches that will give you a greater chance of success:

    1. Since the boundary condition will cause 1 - 2y to be negative, re-write the DE as dt = - \frac{dy}{2y - 1}, integrate in the usual way and then include the boundary condition in the usal way, or

    2. Make y the subject before using the boundary condition.
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  5. #5
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    Quote Originally Posted by latavee View Post
    so the -1/2 ln(u) +C isn't correct? Thanks! I found your tutorials very helpful!
    My understanding is it IS correct, but in complex domain. Now you have obtained
     -\tfrac{1}{2} \ln(1-2y) = t + C (1)

    Put in y(0)=\tfrac{5}{2}, you get
    C=-\tfrac{1}{2}\ln(-4) (2)

    Put (2) into (1), you should get
     -\tfrac{1}{2} \ln(1-2y) = t -\tfrac{1}{2}\ln(-4) (3)

    - After some simplification, you will obtain
     -\tfrac{1}{2} \ln \frac{1-2y}{-4} = -\tfrac{1}{2} \ln \frac{2y-1}{4}= t (4)
    which comes back to real domain.
    Last edited by mr fantastic; September 18th 2009 at 08:00 AM. Reason: Restored original post
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