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Math Help - Solving a DE...need help!

  1. #1
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    Solving a DE...need help!

    Hey.

    My problem is y(t)' + (1/y(t)+1) = 0, where y(0) =2.

    I am having a hard time getting y(t).

    At first glance, this looks like a homogenous ODE...but using the form y'(t) + p(t)y(t)=0, what would p(t) be?

    After hitting that dead end, I decided to try moving the (1/y(t)+1) to the right side and doing
    y'(t) = -(1/y(t)+1) then
    dy/dt = -(1/y(t)+1) then
    dy = -(1/y(t)+1) dt and acting like -(1/y(t)+1) is a constant.

    When I try to do this, I end up with y = (-t/(y(t)+1)) + C
    which leaves me with y(t)^2 + y(t) = t...and I get stuck.

    Also, I have tried getting (y(t)+2)/(y(t) + 1) on the left side by making the left side one fraction and moving the dt over to the right...after I integrate, I get ln(|y(t)+1|)+y(t)=C...and I get stuck.

    Am I forgetting some algebra here, or am I completely lost?

    Please help me!

    Thanks!
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  2. #2
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    Quote Originally Posted by kpizle View Post
    Hey.

    My problem is y(t)' + (1/y(t)+1) = 0, where y(0) =2.

    I am having a hard time getting y(t).

    At first glance, this looks like a homogenous ODE...but using the form y'(t) + p(t)y(t)=0, what would p(t) be?

    After hitting that dead end, I decided to try moving the (1/y(t)+1) to the right side and doing
    y'(t) = -(1/y(t)+1) then
    dy/dt = -(1/y(t)+1) then
    dy = -(1/y(t)+1) dt and acting like -(1/y(t)+1) is a constant.

    When I try to do this, I end up with y = (-t/(y(t)+1)) + C
    which leaves me with y(t)^2 + y(t) = t...and I get stuck.

    Also, I have tried getting (y(t)+2)/(y(t) + 1) on the left side by making the left side one fraction and moving the dt over to the right...after I integrate, I get ln(|y(t)+1|)+y(t)=C...and I get stuck.

    Am I forgetting some algebra here, or am I completely lost?

    Please help me!

    Thanks!
    Is the DE

    y'(t) + \frac{1}{y(t) + 1} = 0

    or

    y'(t) + \frac{1}{y(t)} + 1 = 0?
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Is the DE

    y'(t) + \frac{1}{y(t) + 1} = 0

    or

    y'(t) + \frac{1}{y(t)} + 1 = 0?

    It's y'(t) + \frac{1}{y(t) + 1} = 0

    Sorry about the confusion.
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  4. #4
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    Quote Originally Posted by kpizle View Post
    It's y'(t) + \frac{1}{y(t) + 1} = 0

    Sorry about the confusion.
    y'(t) + \frac{1}{y(t) + 1} = 0

    \frac{dy}{dt} = -\frac{1}{y + 1}

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

    \frac{dt}{dy} = -y - 1

    t = \int{-y - 1\,dy}

    t = -\frac{1}{2}y^2 - y + C.


    You also know y(0) = 2...

    0 = -\frac{1}{2}(2)^2 - 2 + C

    0 = -2 - 2 + C

    0 = -4 + C

    C = 4.


    So the solution is

    t = -\frac{1}{2}y^2 - y + 4

    -2t = y^2 + 2y - 8

    -2t = y^2 + 2y + 1^2 - 1^2 - 8

    -2t = (y + 1)^2 - 9

    9 - 2t = (y + 1)^2

    y + 1 = \pm \sqrt{9 - 2t}

    y = -1 \pm \sqrt{9 - 2t}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    y'(t) + \frac{1}{y(t) + 1} = 0

    \frac{dy}{dt} = -\frac{1}{y + 1}

    \frac{dt}{dy} = -(y + 1) RIGHT HERE, I MADE A WRONG TURN

    \frac{dt}{dy} = -y - 1

    t = \int{-y - 1\,dy}

    t = -\frac{1}{2}y^2 - y + C.


    You also know y(0) = 2...

    0 = -\frac{1}{2}(2)^2 - 2 + C

    0 = -2 - 2 + C

    0 = -4 + C

    C = 4.


    So the solution is

    t = -\frac{1}{2}y^2 - y + 4

    -2t = y^2 + 2y - 8

    -2t = y^2 + 2y + 1^2 - 1^2 - 8

    -2t = (y + 1)^2 - 9

    9 - 2t = (y + 1)^2

    y + 1 = \pm \sqrt{9 - 2t}

    y = -1 \pm \sqrt{9 - 2t}.

    Oh my gosh!!

    You flipped everything!!!! Of course!

    I have been beating this thing for about two hours now and thought I had exhausted all the tricks.

    My goodness, that simplies things.

    Thank you so very much! I really appreciate it.
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  6. #6
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    Quote Originally Posted by kpizle View Post
    Oh my gosh!!

    You flipped everything!!!! Of course!

    I have been beating this thing for about two hours now and thought I had exhausted all the tricks.

    My goodness, that simplies things.

    Thank you so very much! I really appreciate it.
    Yes, I like the flipping method.

    Another alternative is to separate the variables.

    \frac{dy}{dt} = -\frac{1}{y + 1}

    (y + 1)\,\frac{dy}{dt} = -1

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

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


    You should get the same answer.
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