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Find all solutions?

  1. Oct 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of xy'=2-x+(2x-2)y-xy^2.

    2. Relevant equations
    None.

    3. The attempt at a solution
    The answer in the book is y=1-1/(x(1-cx)).
    Here's my work:
    xy'=2-x+2xy-2y-xy^2
    xy'=(xy^2-2xy+x)+2(1-y)
    xy'=x(y-1)^2-2(y-1)
    y'=(y-1)^2-2(y-1)/x
    u=y-1
    u'=y'
    u'=u^2-2u/x
    And now I'm stucked. Please help me, I've struggled with this problem for a long time.
     
  2. jcsd
  3. Oct 30, 2014 #2

    Mark44

    Staff: Mentor

    You lost a minus sign. In the equation before the one above, you have -xy2 that became xy2.
    I don't have any more suggestions at the moment, but I'll take a closer look in a little while.

    BTW, there is no such word in English as "stucked."
     
  4. Oct 30, 2014 #3

    HallsofIvy

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    You lost more than just the negative on "[itex]-xy^2[/itex]"! You should have
    [tex]xy'= -(xy^2- 2yx+ x)+ 2(1- y)[/tex]

     
  5. Oct 30, 2014 #4

    Mark44

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    Math10,
    After you fix your algebra errors, with your substitution you should have an equation in x, u, and u'. Try the substitution w = ux. That should get you a new DE that is separable.
     
  6. Oct 30, 2014 #5
    Let me try.
     
  7. Oct 30, 2014 #6
    Here's my work:

    xy'=2-x+2xy-2y-xy^2
    xy'=-(xy^2-2xy+x)+2(1-y)
    xy'=-x(y-1)^2-2(y-1)
    y'=-(y-1)^2-2(y-1)/x
    u=y-1
    u'=y'
    u'=-u^2-2u/x
    w=ux
    u=w/x
    u'=w'
    w'=-(w^2/x^2)-(2w/x)(1/x)
    w'=-w^2/x^2-2w/x^2
    w'=(-w^2-2w)/x^2
    dw/(-w^2-2w)=dx/x^2
    dw/(-w(w+2))=dx/x^2
    A/-w+B/(w+2)=1
    A(w+2)-Bw=1
    A=1/2, B=1/2
    (-1/2)ln abs(w)+(1/2)ln abs(w+2)=-1/x+C
    (-1/2)(ln abs(w)-ln abs(w+2))=-1/x+C
    ln abs(w/(w+2))=2/x+C
    w/(w+2)=Ce^(2/x)
    w=Ce^(2/x)(w+2)
    w=wCe^(2/x)+2Ce^(2/x)
    w-wCe^(2/x)=2Ce^(2/x)
    w(1-Ce^(2/x))=2Ce^(2/x)
    w=2Ce^(2/x)/(1-Ce^(2/x))
    u=2Ce^(2/x)/(x(1-Ce^(2/x)))
    y=1+2Ce^(2/x)/(x(1-Ce^(2/x)))
    This doesn't match the answer in my book. Please check my work carefully and tell me where I got wrong if I'm incorrect.
     
  8. Oct 30, 2014 #7

    haruspex

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    You didn't mean that, right?
     
  9. Oct 31, 2014 #8
    So where did I make a mistake? What's wrong?
     
  10. Oct 31, 2014 #9

    Mark44

    Staff: Mentor

    You're treating x as if it were a constant - it's not. Instead of working with the equation, use the equivalent equation w = vx. Now, what is w'? What differentiation rule should come to mind?
     
  11. Oct 31, 2014 #10
    Let me try.
     
  12. Oct 31, 2014 #11
    I see how I got it wrong.
    So with the substitution w=ux,
    u=w/x
    u'=(xw'-wx')/x^2
    (xw'-wx')/x^2=-(w^2/x^2)-(2w/x)(1/x)
    (xw'-wx')/x^2=(-w^2-2w)/x^2
    Now what should I do?
     
  13. Oct 31, 2014 #12

    haruspex

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    and x' equals what?
    There is a fairly obvious cancellation.
     
  14. Oct 31, 2014 #13
    So I got
    xw'-wx'=-w^2-2w
    wx'=xw'+w^2+2w
    x'=(xw'+w^2+2w)/w
    x'=w'/u+w+2
    Now what?
     
  15. Oct 31, 2014 #14

    haruspex

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    I ask again, what is x' equal to? Think about what it means.
    There's no point in reintroducing u.
     
  16. Oct 31, 2014 #15
    x'=w'x/w+w+2
    Isn't it?
     
  17. Oct 31, 2014 #16

    Mark44

    Staff: Mentor

    Why did you make it harder on yourself by solving for u again? It's a lot simpler to differentiate the equation above than the one you have below.

    And think about what you're doing. You're differentiating with respect to which variable?
     
  18. Oct 31, 2014 #17

    Mark44

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    No.
    What does x' mean?
     
  19. Oct 31, 2014 #18

    RUber

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    The assumption for the beginning was that y was a function of x.
    y'=dy/dx.
     
  20. Nov 1, 2014 #19
    So w'=ux'+xu' and what do I do next?
     
  21. Nov 1, 2014 #20

    Mark44

    Staff: Mentor

    What is x'?

    By that, I mean what does x' mean?
     
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