1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equation problem

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

    x^2dx + y(x-1)dy = 0

    I obtained a solution of

    \frac{1}{2}x^2 + x + ln | x-1 | + \frac{1}{2}y^2 = C

    The book, however, gives an answer of


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


    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 :)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differential Equation problem