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Math Help - 1st order diff equ-separation of variables EASY!!!

  1. #1
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    1st order diff equ-separation of variables EASY!!!

    Okay lets see if someone on here can help me. This problem looks pretty easy:
    dy/dt=1(1-y)
    I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
    t+c=ln(y)-ln(y-1)+c
    now I raise each side exponentially to get rid of the ln's:
    e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
    this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
    y(t)=ce^t/(ce^t+1)
    I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mxrider777 View Post
    Okay lets see if someone on here can help me. This problem looks pretty easy:
    dy/dt=1(1-y)
    I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
    t+c=ln(y)-ln(y-1)+c
    now I raise each side exponentially to get rid of the ln's:
    e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
    this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
    y(t)=ce^t/(ce^t+1)
    I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?
    \frac{1}{1-y}dy=(1)dt

    \int\frac{1}{1-y}dy=\int(1)dt.

    Can you finish?

    EDIT: That wasn't youre trouble. Sorry, misread.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mxrider777 View Post
    Okay lets see if someone on here can help me. This problem looks pretty easy:
    dy/dt=1(1-y)
    I see that its an autonomous equation that will allow for separation of variables, which I do. I separate and integrate accordingly which yields me:
    t+c=ln(y)-ln(y-1)+c
    now I raise each side exponentially to get rid of the ln's:
    e^(t+c)=e^(ln(y))-e^(ln(y-1))+e^c
    this is where I think I am making a mistake. There must be some law of logarithms that I don't see because the answer turns out to be (according to the back of my book):
    y(t)=ce^t/(ce^t+1)
    I can only come up with the ce^t and the rest is wrong. Can someone fill me in to what I need to do/what I am doing wrong?
    check what you wrote. the solution you claim the book has is not the solution to your problem. the solution to \frac {dy}{dt} = 1 - y is y = 1 - Ce^{-t} NOT y = \frac {Ce^t}{Ce^t + 1}
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  4. #4
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    just for information:

    y=\frac{Ce^t}{Ce^t+1} is the solution of \frac{dy}{dt}=y(1-y).

    Maybe a typo in the book?

    Coomast
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