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Solving a separable ODE.

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data
    solve y(xy+1)dx + x(1+x^2y^2)dy=0

    3. The attempt at a solution

    well, I substituted u=xy. Here is what I've done so far.
    du = xdy + ydx -> xdy= du - ydx -> xdy = du - (u/x)dx


    (u/x)(u+1)dx + x(1+u^2)dy=0
    (u/x)(u+1)dx + (1 + u^2)(du - (u/x)dx)=0
    (u^2/x)dx + (u/x)dx + du - (u/x)dx + (u^2)du - (u^3/x)dx = 0
    (u^2)dx + xdu + (u^2x)du - (u^3)dx = 0

    dx/x + (1+u^2)/(u^2 - u^3)du = 0

    hence, dx/x = (1+u^2)/(u^3 - u^2)du, if we integrate it we'll have:

    ln|x| = -ln|u| + ln|u+1| + ln|u-1| + C
    ln|x| = ln|(u^2 - 1)/u| + lnC'

    x = C(u^2 -1)/u -> x = C(xy - 1/(xy)).

    but the book says that the final answer is ln|(1-xy)^2/y| + 1/(xy) = C. :|

    Have I differentiated u correctly? I can't find out why my answer is incorrect.
     
  2. jcsd
  3. Oct 20, 2011 #2

    vela

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    Seems you made an algebra error somewhere. According to Mathematica, you should get
    [tex]\frac{1+u^2}{u^3-u^2} = \frac{2}{u-1}-\frac{1}{u}-\frac{1}{u^2}[/tex]
     
  4. Oct 20, 2011 #3

    HallsofIvy

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    You seem to have factored [itex]x^3- x^2[/itex] as [itex]x(x- 1)(x+ 1)[/itex]. That is incorrect: [itex]x(x- 1)(x+ 1)= x^3- x[/itex]. Rather, factor out the full "[itex]x^2[/itex]" to get [itex]x^2(x- 1)[/itex].
     
  5. Oct 20, 2011 #4
    Thanks. but there is still a little problem. now I have:

    ln|x| = 1/u - ln|u| + 2ln|1-u| + C -> ln|x| = 1/(xy) - ln|xy| + ln|(1-xy)^2| + C
    ln|x| = 1/(xy) - ln|x| +ln|((1-xy)^2)/y| + C.

    but I need to cancel out ln|x|. I'm sorry that I'm asking you to find my algebra errors, it's because I lack concentration now and when I make a naive algebra error I can't find it myself even if I double check everything I've done from the beginning.
     
  6. Oct 20, 2011 #5

    vela

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    I don't see anything wrong in your work. It could be the book's wrong or you copied the original problem down wrong.
     
  7. Oct 20, 2011 #6

    SammyS

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    ln(|xy|) = ln(|x||y|) = ln(|x|) + ln(|y|)

    Then subtract ln(|x|) from both sides.
     
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