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Two Questions (antiderivative and diff eq.)

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    (i've put my questions in bold)

    The differential equation is

    dy/dx = xy^2

    I separated variables and came up with (solving for c)

    c = -1/y - (x^2)/2

    I then had to find the particular solution for y(0) = 1, so I came up with

    c = -1 - 0 so c = -1

    I then solved for y and got that y = -1/(1-(x^2)/2) ... is this the correct answer for the specific solution?

    Finally, I had to determine what values of C gave the diff eq. vertical asymptotes. I know this happens when the denominator equals zero, so I said that it is the case when

    c+(x^2)/2 = 0. This only happens when C = 0 and x = 0. Correct?

    1. The problem statement, all variables and given/known data

    The second question:

    the function G(x) = G(-4) + antiderivative from -4 to x of f(t) dt over the interval

    I have the graph of f, however, so when I take G'(x) I get f(x), correct? What about the G(-2)? That's the part I am not sure of. I'm assuming that it doesn't matter since the interval runs from [-4,3] and the antiderivative from -4 to -4 is simply zero. Is this the right way of looking at it?

  2. jcsd
  3. Mar 12, 2008 #2


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    You can check it youself. Does y satisfy the original differential equation? Does it satisfy y(0)= -1?

    Suppose c= -2? What would happen at x= 2 or -2? Are you told to assume that c cannot be negative?

    What does "from -4 to -4" have to do with anything" [itex]G(-2)= G(-4)+ \int_{-4}^{-2}f(x)dx[/itex]. But there was no mention of "G(-2)" in the statement of the question. Did you mean G(-4)? In fact, there is no question in the statement of the question! There is simply a definition of the function G(x). What is the question? To find the derivative of G? If so, then G(-4) is a constant and the derivative of a constant is 0.
  4. Mar 12, 2008 #3
    Thanks very much for your help. I apologize if some of the questions were dumb / elementary...but I've figured out how I was thinking wrongly about the problem.
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