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Math Help - Ordinary Differential Equations - tough IVP

  1. #1
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    Exclamation Ordinary Differential Equations - tough IVP

    Hi, I am supposed to solve the following initial value problem but am stuck with an ugly integral:

    y' = y(1-y) y(0)=yo (y subscript 0), where is an arbitrary constant.

    I have worked it out as follows:


    dy/dt = (y-y)

    dy/(y-y) = dt

    [ dy/(y-y) ] = [ dt ]

    [ dy/(y-y) ] = t + c


    But how do you integrate the left side? Or is there another way to solve this for y? Thank you.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by thecelticswin View Post
    Hi, I am supposed to solve the following initial value problem but am stuck with an ugly integral:

    y' = y(1-y) y(0)=yo (y subscript 0), where is an arbitrary constant.

    I have worked it out as follows:


    dy/dt = (y-y)

    dy/(y-y) = dt

    [ dy/(y-y) ] = [ dt ]

    [ dy/(y-y) ] = t + c


    But how do you integrate the left side? Or is there another way to solve this for y? Thank you.
    \int\frac{\,dy}{y-y^2}=\int\frac{\,dy}{y(1-y)}

    You can apply partial fraction decomposition

    -OR-

    \frac{1}{y(1-y)}=\frac{(1-y)+y}{y(1-y)}=\frac{1-y}{y(1-y)}+\frac{y}{y(1-y)}=\frac{1}{y}-\frac{1}{y-1}

    So, \int\frac{\,dy}{y(1-y)}=\int\left[\frac{1}{y}-\frac{1}{y-1}\right]\,dy

    This is a bit easier to integrate now.

    Does this make sense?

    --Chris
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  3. #3
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    thank you, this certainly does make it easier.
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  4. #4
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    another problem...

    I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

    y= -e^(
    t+c) / (1-e^(t+c))


    however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
    t)],
    where y(0) = yo. How did this result get obtained from the integral??
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  5. #5
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    Quote Originally Posted by thecelticswin View Post
    I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

    y= -e^(
    t+c) / (1-e^(t+c))


    however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
    t)],
    where y(0) = yo. How did this result get obtained from the integral??
    Substitute the initial condition. In other words, when x=0, y=y_0. Solve for the constant.
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    I understand this, but still cannot get my equation to look like the answer. Could you provide any assistance?
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  7. #7
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    or is e^c = yo??
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  8. #8
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    Quote Originally Posted by thecelticswin View Post
    I seem to have run into another snag. I solved the integral and have: ln|y|-ln|y-1|=t+c. Solving for y, I obtained:

    y= -e^(
    t+c) / (1-e^(t+c))


    however, the answer that I am supposed to find is y=yo/[yo+(1-yoe^-
    t)],
    where y(0) = yo. How did this result get obtained from the integral??
    I just worked out the integral again, and received a different answer...

    \ln |y| - \ln |y-1| = \alpha t + c
    \ln \left|\frac{y}{y-1}\right | = \alpha t + c
    \left | \frac{y}{y-1} \right | = e^{\alpha t + c}
    \left | \frac{y}{y-1} \right | = e^c e^{\alpha t}
    \frac{y}{y-1} = \pm e^c e^{\alpha t}
    \frac{y}{y-1} = A e^{\alpha t}
    1 + \frac{1}{y-1} = A e^{\alpha t}

    Can you go from here?
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  9. #9
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    Err, for some reason the y and yo are confusing me. From this point, I did:

    1 / (y-1) = -1+Ae^t

    y-1 = 1 / (-1+Ae^
    t)

    y = (1 + Ae^
    t - 1) / (Ae^t - 1)

    y = (
    Ae^t ) / (Ae^t - 1)

    when t=0, y=yo

    yo = A/(A-1)

    yoA-yo = A

    -yo = A(1-yo)

    A = (1-yo)/-yo = (yo-1)/yo

    And now I am stuck again. When I plug this in for A, I still cannot get it to look right..
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  10. #10
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    Quote Originally Posted by thecelticswin View Post
    Err, for some reason the y and yo are confusing me. From this point, I did:

    1 / (y-1) = -1+Ae^t

    y-1 = 1 / (-1+Ae^
    t)

    y = (1 + Ae^
    t - 1) / (Ae^t - 1)

    y = (
    Ae^t ) / (Ae^t - 1)

    when t=0, y=yo

    yo = A/(A-1)

    yoA-yo = A

    -yo = A(1-yo)

    A = (1-yo)/-yo = (yo-1)/yo

    And now I am stuck again. When I plug this in for A, I still cannot get it to look right..
    That's because A = \frac{y_0}{y_0 - 1}, not \frac{y_0 - 1}{y_0}
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  11. #11
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    Ok, got it. Thank you everyone.
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