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Separable equations-check my answers please

  1. Nov 7, 2006 #1

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

    1. Give the general equation of the curves that satisfy this equation.

    2. Show that the straight lines y=1 and y=-1 are also solutions

    3. Do any of the curves you found in 1) intersect y=1?

    My Ans:
    1. The general solution i found out to be x^2 + C = (y-1)/(y+1), correct?

    2. Does that mean y=1 is a solution and y=-1 is not because it will have a denominator of zero?

    3. The only intersection is at x=0, y=1 , since x can't be 0, there are no curves intersecting y=1

    Am i doing anything wrong here?
     
  2. jcsd
  3. Nov 8, 2006 #2

    dextercioby

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    y=-1 is a valid solution as you can see from the ODE. y=1 as well. For the solution curves they are horizontal asymptotes.

    Daniel.
     
  4. Nov 8, 2006 #3

    HallsofIvy

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    2x= {1(y+1)-1(y-1)}/(y+1)^2 dy/dx= 2/(y+1)^2 dy/dx so
    dy/dx= x(y+1)^2. Doesn't look anything like dy/dx= (y^2-1)/x to me!
    I assume you got dy/(y^2-1)= dx/x. How did you integrate that? Be careful with your constant of integration. If ln a= ln b+ C then a= eCb= C'b where C'= eC. The point is that the constant is now multiplying not added to b.

    No, "Show that the straight lines y=1 and y=-1 are also solutions" implies that both y= 1 and y= -1 are solutions! There is no "denominator of zero" because the equation as given, dy/dx= (y^2 -1)/x, has only x as denominator. y= 1 or y= -1 can't make the denominator 0. If you are referring to the y+1 in the denominator of your general solutions, the whole point of this exercise is to show you that there may be solutions to a d.e. that your "general" solution from integrating does not give.
    Suppose y= 1 for all x. What is dy/dx? What is (y^2- 1)/x when y= 1? Same questions for y= -1.

    Recalculate your "general solution" and try this again. In fact, none of the solutions intersect y= 1 but not because of any problem at x= 0.
     
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