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Homework Help: 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

    Staff: Mentor

    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

    Staff: Mentor

    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?
     
  22. Nov 1, 2014 #21

    Mark44

    Staff: Mentor

    Your work above is what we call a "wall of text." I don't see a single space anywhere in your equations. This makes it much more difficult to understand what you have written. You can make things more readable by inserting spaces around addition and subtraction operations, and around '='.

    Here's one of your equation with some spaces added:
    w = wCe^(2/x) + 2Ce^(2/x)
    If you want us to help you, put some extra effort into making it easier for us to do so.
     
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