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Homework Help: Need Help With a Separable Differential Equation

  1. Sep 6, 2014 #1
    Hello. I need some help solving a differential equation. I think where I'm going wrong is integrating one side via partial fractions, but I'm not quite sure where my mistake is. Using Wolfram, I found the correct answer, which is below. Thanks.

    1. The problem statement, all variables and given/known data
    Solve the following initial value problem:

    [itex]ty' = y^2 - y[/itex]
    [itex]y(1) = \frac{1}{2}[/itex]
    [itex]y = ?[/itex]

    2. Relevant equations


    3. The attempt at a solution

    [itex]\int \frac{dy}{y(y-1)}\ [/itex] Eq. 1

    [itex]\frac{1}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1} [/itex]

    [itex]A(y-1) + By = 1 [/itex]
    [itex]y = 1 -> B = 1 [/itex]
    [itex]y = 0 -> A = -1 [/itex]

    [itex]\int -\frac{1}{y} + \frac{1}{y-1}dy = ln|y-1|-ln|y| + c = ln|\frac{y-1}{y}| + c[/itex]

    However, Wolfram tells me that the answer is actually:

    [itex]ln|\frac{1-y}{y}| + c[/itex]

    If I go on to use that in the problem, I get the correct answer for the DE.

    [itex]y = \frac{1}{t + 1} (c = 0)[/itex]

    With my answer, I get the incorrect overall answer of:

    [itex]y = \frac{1}{1 - t}[/itex]

    How am I solving the partial fractions problem incorrectly?

  2. jcsd
  3. Sep 6, 2014 #2
    ##|y-1|=|(-1)(1-y)|=|-1||1-y|=|1-y|##, so your intermediate answer is equivalent to Wolfram's.

    Without seeing your work, it's impossible to know why your end answer is wrong. But I'm guessing it has something to do with not dealing with the absolute value properly. Or maybe it's just some lazy algebra, considering your answer differs from the correct one by a sign. I would point out that [itex]y = \frac{1}{1 - t}[/itex] isn't even defined at ##t=1##. So something definitely went wrong somewhere.
  4. Sep 7, 2014 #3
    Good point that my answer isn't even defined there.

    I end up with:

    [itex]ln|t| = ln|\frac{y-1}{y}| + c[/itex]

    c ends up being 0 with the initial value, so that simplifies to:

    [itex]|t| = |\frac{y-1}{y}|[/itex]

    I suppose here is where I need to look at t equal to the positive or negative value of the entire quantity in the right side absolute value.

    [itex]t = \frac{y-1}{y}[/itex]
    [itex]t = -(\frac{y-1}{y})[/itex]

    The second one gives me the correct answer. I guess each time I do this, I just need to make sure that the function is defined at the given condition? There should be no instance where the same initial value has two solutions, right?

    Thanks again.
  5. Sep 7, 2014 #4
    There are a few ways to handle this one. My personal favorite is to realize that the original "intermediate" solution is equivalent to $$\ln|Ct|=\ln\left|\frac{y-1}{y}\right|$$. All of the silliness with the absolute values gets "absorbed" by the undetermined constant of integration ##C##, and we get $$Ct=\frac{y-1}{y}$$ Now we can use the initial condition to solve for ##C## and then put ##y## in terms of ##t##. Or do it the other way around. It really doesn't matter at this point.
  6. Sep 7, 2014 #5
    That's a neat trick. So basically, my equation

    [itex]ln|t| = ln|\frac{y-1}{y}| + c[/itex]

    turns into

    [itex]ln|t| + b = ln|\frac{y-1}{y}| [/itex] with b = -c

    Then ln|t| + b can be thought of as ln|t| + ln|C|, which is then ln|Ct|. If I'm getting the logic right.

    Nice. Thanks for the tip.
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