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Math Help - Elementary Autonomous Initial Value Textbook Problem

  1. #1
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    Elementary Autonomous Initial Value Textbook Problem

    Solve the following initial value problem and determine the interval where the solution exists.

    dy/dt = 2ty^2
    y(0) = y_0

    Textbook answer is:

    When y_0 <> 0, then y = 1/(1/y_0 - t^2).
    When y_0 = 0, then y = 0

    When y_0 > 0, the interval where solution exists is |t| < 1/sqrt(y_0)
    When y_0 <= 0, the interval where solution exists is -infinity < t < +infinity


    My work:

    The equation is separable:

    dy/y^2 = 2t dt
    -1/y = t^2 + C

    When y_0 = 0, I use intuition to come to the same solution that the textbook shows.

    When y_0 <> 0
    C = -1/y_0
    y = 1/(1/y_0 - t^2).
    The interval where the solution exists is where the denominator expression does not equal zero. Algebraically, that simplifies to t <> 1/sqrt(y_0)

    The textbook answer is different. What did I do wrong or omit?
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  2. #2
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    So far so good, you have
    y=\frac{1}{\frac{1}{y_0}-t^2}.
    Now rearrange this a litlle by multiplying the top and the bottom by y_0 to get
    y=\frac{y_0}{1-y_0t^2}.
    Now your correct in saying that this exists provided the denominator is non-zero, or when y_0t^2 \not = 1. But note that t^2 is always greater than zero, hence when y_0 is negative, it is impossible for t^2y_0 to be positive, hence the equation exists for all y_0<0, t. For y_0>0 the denominator is equal to zero when t=+-\frac{1}{\sqrt{y_0}}.

    That agree with your book?
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  3. #3
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    Quote Originally Posted by Diemo View Post
    So far so good, you have
    y=\frac{1}{\frac{1}{y_0}-t^2}.
    Now rearrange this a litlle by multiplying the top and the bottom by y_0 to get
    y=\frac{y_0}{1-y_0t^2}.
    Now your correct in saying that this exists provided the denominator is non-zero, or when y_0t^2 \not = 1. But note that t^2 is always greater than zero, hence when y_0 is negative, it is impossible for t^2y_0 to be positive, hence the equation exists for all y_0<0, t. For y_0>0 the denominator is equal to zero when t=+-\frac{1}{\sqrt{y_0}}.

    That agree with your book?
    When
    y_0 > 0

    then your work and my work indicate that the solution exists for
    t \ne \pm \frac{1}{\sqrt{y_0}}

    the book says that the solution exists for
    |t| < \frac{1}{\sqrt{y_0}}

    If I set y_0 = 1 and graph the corresponding formula y = \frac{1}{1-t^2}, the graph does not exist at t = \pm 1, but exists at all other values of t.

    It would seem that the book is simply wrong? For one question I wouldn't be surprised (I find at least one or two mistakes in every text book), but there were three questions in a row that were differing from my answers like this and that usually means I am making the mistake rather than the book.
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