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Homework Help: Differential Equation problem

  1. Nov 20, 2004 #1
    I'm asked to find the general solution of the differential equation

    [tex]
    x^2dx + y(x-1)dy = 0
    [/tex]

    I obtained a solution of

    [tex]
    \frac{1}{2}x^2 + x + ln | x-1 | + \frac{1}{2}y^2 = C
    [/tex]

    The book, however, gives an answer of

    [tex]

    (x+1)^2 + y^2 + 2ln |C(x-1)| = 0

    [/tex]

    I'm sure it's a simplified answer of my own answer. What I don't understand is how a term of +1 and +ln c appeared in the equation after transposing it in the equation. I know for a fact that you can transofrm c into (ln c) since they are constants, but the rest I dont get.

    Thanks in advance.
     
  2. jcsd
  3. Nov 20, 2004 #2
    Well, first of all you can multiply your equation with 2, and rewriting you constant to -(2Ln[C] +1) (which is ofcourse just another constant). After that the other answer more or less rolls out:

    [tex]x^2 + 2x + 2ln | x-1 | + y^2 = -(2ln|C| + 1)[/tex]
    [tex]x^2 + 2x + 1 + y^2 + 2 ln | x-1 | + 2ln|C| = 0[/tex]
    [tex](x+1)^2 + y^2+ 2 ln|C(x-1)| = 0[/tex]
     
  4. Nov 20, 2004 #3
    i see. I never knew you could add numbers to C. Thanx :)
     
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