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Solving First Order Linear Differential Equation

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Given (E): [itex](x+1)^{2}(xy'-y) = -(2x+1)[/itex]

    Determine the set of applications from the interval [itex]I[/itex] to [itex]ℝ[/itex] which are solutions of (E) for:

    a) [itex]I = (0,+∞)[/itex]
    b) [itex]I = (-1,0)[/itex]
    c) [itex]I = (-∞,-1)[/itex]
    d) [itex]I = (-1,+∞)[/itex]
    e) [itex]I = ℝ[/itex]


    The attempt at a solution

    I have rearranged (E) into the form:

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

    and chosen integrating factor:

    [itex]exp(∫\frac{-1}{x}dx) = \frac{1}{x}[/itex]

    now multiplying by the integrating factor and integrating gives:

    [itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-(2x+1)}{x^{2}(x+1)^{2}}dx[/itex]

    partial fractions now give:

    [itex]∫\frac{y'}{x}-\frac{y}{x^{2}}dx = ∫\frac{-1}{x^{2}}+\frac{1}{(x+1)^{2}}dx[/itex]

    After integrating and a tiny bit of algebra I get a general solution, for [itex]k\inℝ[/itex] as:

    [itex]y=\frac{1}{x+1}+kx[/itex]

    Now I think I am right in saying that this solution works fine for the intervals in a), b) and c)?

    However for [itex]I = (-1,+∞)[/itex] there is (or potentially is) a problem with the point [itex]0[/itex] and for [itex]I = ℝ[/itex] there are (or potentially are) problems with the points [itex]0[/itex] and [itex]-1[/itex]?

    Now I'm not sure how to proceed. Am I thinking along the right lines?

    Any help/pointers would be very much appreciated, Thanks!
     
  2. jcsd
  3. Apr 23, 2012 #2
    Can the general solution already posted simply extend to [itex]I = (-1,+∞)[/itex] ??
    If so how can this be shown in detail?

    Also I don't yet understand what happens when we are considering [itex]ℝ[/itex]

    Thanks for any help!
     
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