# Implicif differential equation and zero

1. Oct 25, 2005

### Yann

I have an implicit differential equation;

$$\left(x^2+y^2\right)^2=x^2-y^2$$

With a little work;

$$y' = -\frac{x\left(2x^2+2y^2-1\right)}{y\left(2y^2+2x^2+1\right)}$$

Now I have to find the 4 points where y' = 0, how can I do that ?

2. Oct 25, 2005

### mathmike

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

3. Oct 25, 2005

### Pseudo Statistic

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
4. Oct 26, 2005

### HallsofIvy

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

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

Putting that back into the original equation,
$$x^2- y^2= \frac{1}{2}$$
That's a hyperbola that intersects the circle above when (add the two equations) 2x2= 1 or
$$x= \pm\frac{\sqrt{2}}{2}$$
Of course
$$y= \pm\frac{\sqrt{2}}{2}$$
also and those give the four points.

Last edited: Oct 26, 2005
5. Oct 27, 2005

### Yann

Sorry, but i don't understand the "Putting that back into the original equation". Y' = 0 when the circle and the basic equation intersect;

$$(x^2+y^2)^2 - x^2+y^2 = 2x^2+2y^2-1$$

How can you solve that ?

6. Oct 27, 2005

### amcavoy

HallsofIvy just showed you. Expand what you have posted there, and simplify! You'll see that some terms cancel out.

7. Oct 27, 2005

### Yann

How ? I understand the idea, but I cannot solve it. There's two unkown. By expanding I have;

$$x^4+2x^2y^2+y^4-3x^2-y^2 +1 = 0$$

It's not better.

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

$$\left(\frac{1}{2}\right)^2 = x^2-y^2 \to x^2-y^2 = \frac{1}{4}$$

Which make no sence !

Last edited: Oct 27, 2005