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Homework Help: Implicif differential equation and zero

  1. Oct 25, 2005 #1
    I have an implicit differential equation;

    [tex]\left(x^2+y^2\right)^2=x^2-y^2[/tex]

    With a little work;

    [tex]y' = -\frac{x\left(2x^2+2y^2-1\right)}{y\left(2y^2+2x^2+1\right)}[/tex]

    Now I have to find the 4 points where y' = 0, how can I do that ?
     
  2. jcsd
  3. Oct 25, 2005 #2
    you can graph it or you can sub x = y = 0 for two of the roots and then use the quadratic for the other two factors
     
  4. Oct 25, 2005 #3
    Let's look at your original function:
    (x^2 + y^2)^2 = x^2 - y^2
    Let's parametrize, x = cos t, y = sin t, now:
    (cos^2 t + sin^2 t)^2 = cos^2 t - sin^2 t
    cos^2 blabla + sin^2 blabla = 1, right? So...
    1 = cos^2 t - sin^2 t
    cos^2 t - sin^2 t - 1 = 0
    Differentiating...
    -2sin t cos t - 2 sin t cos t = 0
    -2 sin t cos t = 2 sin t cos t
    - sin t cos t = sin t cos t
    Looking at it, you can tell t = 0 is a solution to this equation, and so is t = pi/2...
    Let's look a bit further..... t = pi is a solution too, isn't it? Also, t = 2pi...
    Looking closer, t = 2pi - pi/2 = 3pi/2 would also yield the same answer, wouldn't it?
    And let's think again... if t were negative, it would also yield the same answers, thus we also have t = -pi/2, t = -2pi, t = -3pi/2
    There, we've got 8 solutions now to this parametrized equation..
    Let's go back to your original equation now...
    x = cos t
    y = sin t
    Let's look at our values of t... 0, pi/2, pi and 2pi... let's plug them in..
    x(0) = 1
    y(0) = 0
    (1,0), let's just keep this in mind for now, shall we?
    x(pi/2) = 0
    y(pi/2) = 1
    (0,1)
    x(pi) = -1
    y(pi) = 0
    (-1, 0)
    x(2pi) = 1
    y(2pi) = 0
    (1,0)
    x(-2pi) = 1
    y(-2pi) = 0
    (1,0)
    x(-3pi/2) = 0
    y = 1
    (0,1)
    x(-pi) = -1
    y(-pi) = 0
    (-1, 0)
    x(-pi/2) = 0
    y(-pi/2) = -1
    (0, -1)
    OK, let's look at the solution sets we have here, (-1,0), (0,1), (1,0), (0, -1)
    From your implicitly differentiated formula, let's plug one of them in, the first one..
    As you can see, it yields a 0 in the denominator, we'll rule it out..
    Let's look at (0,1)....
    It gives you 0 in the numerator.... and in the denominator is leaves you with 3... nice.... we want this.
    Write (0,1) down...
    (1,0) would yield a 0 in the denominator as well.... oh...
    (0, -1) works great too.... doesn't leave us with a nasty 0 in the denominator!
    OK, so far we have (0,-1) and (0,1) working... that leaves us to find two more solutions...
    Let's look at the quadratic...
    2x^2 + 2y^2 - 1 = 0
    Hmm, looks familiar; a circle, isn't it? (re-arranging...)
    x^2 + y^2 = 0.5
    Hmm.. interesting... we can now find out when this will equal 0; let's set x = 0..
    y^2 = 0.5
    y = + or - 1/root 2
    Hmm... interesting... let's plug this into the equation..... it gives us 0!
    OK, we found our four points:
    (0, 1)
    (0, -1)
    (0, 1/root2)
    (0, -1/root2)

    I think those are the answers anyway..
    Oh well, atleast I tried...

    disregard me, my logic is very wrong :P
     
    Last edited: Oct 25, 2005
  5. Oct 26, 2005 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Why not try the obvious: set y' equal to 0 and solve!

    [tex]y' = -\frac{x\left(2x^2+2y^2-1\right)}{y\left(2y^2+2x^2+1\right)}= 0[/tex]
    Immediately multiply both sides by that denominator to get:
    [tex]x\left(2x^2+2y^2-1\right)= 0[/tex]
    x= 0 is an obvious solution, but gives complex y. If x is not 0, then we have
    [tex]x^2+ y^2= \frac{1}{2}[/tex].
    Any point on that circle makes y'= 0.

    Putting that back into the original equation,
    [tex]x^2- y^2= \frac{1}{2}[/tex]
    That's a hyperbola that intersects the circle above when (add the two equations) 2x2= 1 or
    [tex]x= \pm\frac{\sqrt{2}}{2}[/tex]
    Of course
    [tex]y= \pm\frac{\sqrt{2}}{2}[/tex]
    also and those give the four points.
     
    Last edited by a moderator: Oct 26, 2005
  6. Oct 27, 2005 #5
    Sorry, but i don't understand the "Putting that back into the original equation". Y' = 0 when the circle and the basic equation intersect;

    [tex](x^2+y^2)^2 - x^2+y^2 = 2x^2+2y^2-1[/tex]

    How can you solve that ?
     
  7. Oct 27, 2005 #6
    HallsofIvy just showed you. Expand what you have posted there, and simplify! :smile: You'll see that some terms cancel out.
     
  8. Oct 27, 2005 #7
    How ? I understand the idea, but I cannot solve it. There's two unkown. By expanding I have;

    [tex]x^4+2x^2y^2+y^4-3x^2-y^2 +1 = 0[/tex]

    It's not better.

    HallsofIvy said that I can get x^2-y^2 = 1/2 but I can only see;

    [tex]\left(\frac{1}{2}\right)^2 = x^2-y^2 \to x^2-y^2 = \frac{1}{4}[/tex]

    Which make no sence !
     
    Last edited: Oct 27, 2005
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