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Solving differential equation(separable)

  1. Feb 20, 2014 #1
    Let y be a solution of the di®erential equation,

    such that


    Then y(1) is:

    My attempt:


    ln|y-1|= ln |x+1| +C

    Problems start to occur from here:

    My method:

    When x=0, y=0. Hence ln0=ln0+C C=0;
    When x=1, ln|y-1|= ln2 + 0;
    Hence |y-1| = 2
    y=3 or y=-1.

    My teacher's method

    ln|y-1|= ln |x+1| +C


    Because y(0)=0 Hence A=-1

    And when x=1, y-1=-2 y=-1

    I am quite confused. My method seems to be legit also, but why I got different answer.
  2. jcsd
  3. Feb 20, 2014 #2


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    Homework Helper

    You of course mean [itex]\ln 1 = \ln 1 + C[/itex] so that [itex]0 = 0 + C[/itex].

    If you look at the ODE you'll see that if [itex]y < 1[/itex] then [itex]dy/dx < 0[/itex] and if [itex]y > 1[/itex] then [itex]dy/dx > 0[/itex] (in both cases this assumes [itex]x \geq 0[/itex]). Thus if [itex]y(0) < 1[/itex] then [itex]y(x) < 1[/itex] for all [itex]x \geq 0[/itex], so you must take [itex]|y - 1| = 1 - y[/itex].
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