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First order differential equation problem.

  1. Nov 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of 2y(x^3+1)dy + 3x^2(1-y^2)dx = 0

    2. Relevant equations



    3. The attempt at a solution

    So I first grouped the terms with dy or dx

    2y/(1-y^2) dy = -3x^2/(x^3 +1) dx


    after integrating both sides and solving, I got

    ln (1-y^2)= -ln(x^3 +1) + c

    and then after simplifying, it becomes 1-y^2= A/(x^3 + 1) and therefore y^2= -A/(x^3+1) + 1.

    The answer according to the book was y^2= 1 + A(x^3 +1). How did they get that??

    Maybe if i could get rid of the negative sign for ln, it might help....but please if u can help me, it would be appreciated. Thanks.
     
  2. jcsd
  3. Nov 13, 2007 #2
    Your first integration is incorrect.

    [tex]\int \frac{2y}{1-y^2} dy [/tex]

    [tex]u = 1-y^2[/tex] then [tex] du = -2y [/tex]

    Yada yada yada, and you should get

    [tex]\int \frac{2y}{1-y^2} dy = -ln(1-y^2)[/tex]

    The problem is that you missed that minus sign. After that the rest follows.
     
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