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Thread: Solving a DE...need help!

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

    Hey.

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

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

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

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

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

    Also, I have tried getting $\displaystyle (y(t)+2)/(y(t) + 1)$ on the left side by making the left side one fraction and moving the dt over to the right...after I integrate, I get $\displaystyle ln(|y(t)+1|)+y(t)=C$...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 $\displaystyle y(t)' + (1/y(t)+1) = 0$, where $\displaystyle y(0) =2$.

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

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

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

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

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

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

    Please help me!

    Thanks!
    Is the DE

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

    or

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

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

    or

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

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

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

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

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

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

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

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

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


    You also know $\displaystyle y(0) = 2$...

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

    $\displaystyle 0 = -2 - 2 + C$

    $\displaystyle 0 = -4 + C$

    $\displaystyle C = 4$.


    So the solution is

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

    $\displaystyle -2t = y^2 + 2y - 8$

    $\displaystyle -2t = y^2 + 2y + 1^2 - 1^2 - 8$

    $\displaystyle -2t = (y + 1)^2 - 9$

    $\displaystyle 9 - 2t = (y + 1)^2$

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

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

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

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

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

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

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


    You also know $\displaystyle y(0) = 2$...

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

    $\displaystyle 0 = -2 - 2 + C$

    $\displaystyle 0 = -4 + C$

    $\displaystyle C = 4$.


    So the solution is

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

    $\displaystyle -2t = y^2 + 2y - 8$

    $\displaystyle -2t = y^2 + 2y + 1^2 - 1^2 - 8$

    $\displaystyle -2t = (y + 1)^2 - 9$

    $\displaystyle 9 - 2t = (y + 1)^2$

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

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

    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.

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

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

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

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


    You should get the same answer.
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