# Simultaneous equations are missing some solutions

1. Apr 11, 2010

### Koldstream

Hey everyone!

1. The problem statement, all variables and given/known data
This is part of a question on finding minima,maxima and saddle points in multi variable functions.

In this section i need to solve the differential equations:

$$y^{3}+3x^{2}y-y$$ [1]
$$x^{3}+3y^{2}x-x$$ [2]

2. Relevant equations

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3. The attempt at a solution

i divide [1] through by y and [2] through by x.

In doing so i am pretty sure i am losing a solution, but i don't know how to solve it in any other way.

Doing this however is get:

x=0.5 y=0.5
x=0.5 y=-0.5
x=-0.5 y=0.5
x=-0.5 y=-0.5

The book says these are correct however it also mentions these solutions:

x=0 y=0
x=0 y=1
x=0 y=-1
x=1 y=0
x=-1 y=0

I suspect i am losing these by dividing through by x and y, but i don't know any other way to solve it.

Could someone show me how to get the other solutions?

2. Apr 11, 2010

### epenguin

As you have written them these do not look like differential equations, for that matter they are not even equations.

The answer to whatever the problem is, which is not clear, probably involves the symmetry whereby the x in the first equation becomes the y in the second and vice versa.

3. Apr 11, 2010

### Koldstream

sorry i am tired

i mean:

$$y^{3}+3x^{2}y-y=0$$

$$x^{3}+3y^{2}x-x=0$$

They are not supposed to be differential equations this is just part of the problem.

4. Apr 11, 2010

### willem2

if you divide $y^3 + 3x^2y + y = 0$ by y you really factor it into

$y (y^2 + 3x^2 + 1) = 0$ this is true if

$y^2 + 3x^2 + 1 = 0$ or y = 0

5. Apr 11, 2010

### Koldstream

then why does this method not yield all the solutions :(

6. Apr 11, 2010

It does yield all the solutions. If you combine y = 0 with $2^3 + 3y^2 - 1 = 0 [itex] and also x = 0 with [itex] y^2 + 3x^2 - 1 = 0 [itex] and also x = 0 with y=0 you get all the additional solutions 7. Apr 11, 2010 ### Koldstream Thanks Ok i think i see that now. But say we factorise $$x^{3}+3y^{2}x-x=0$$ and get x=0 If we then then plug that result (x=0) into $$x^{3}+3y^{2}x-x=0$$ we get $$0^{3}+3y^{2}\times0-0=0$$ which becomes $$y^{2}\times0=0$$ this means the value of y could be anything! How do we know it is one. (if you put x=0 into $$y^{3}+3x^{2}y-y=0$$ it does work) 8. Apr 12, 2010 ### willem2 Logic if you have ab = 0 then a = 0 OR b=0. Concluding that a = 0 AND b=0 isn't allowed. If you have two equations ab = 0 AND cd = 0 you have (a=0 OR b=0) AND (c=0 OR d=0) this is equivalent to (a=0 AND c=0) OR (a=0 AND d=0) OR (b=0 AND c=0) OR (b=0 AND d=0) 9. Apr 12, 2010 ### epenguin You can and probably should slog out [itex] (y^2 + 3x^2 - 1) = 0$ ...(3)

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

by subtracting one from the other. Then you can factorise the result, then you can see as well as what we have already had, a relation between x and y, which you can substitute to have an equation in only one of them. But when you get the final answer will see what I said above - a symmetry comes into it. Which is, I think, what they wanted you to notice in the first place, and you can perhaps work out, well I don't know that you can in a short time without being shown it once:

$y (y^2 + 3x^2 - 1) = 0$ ...(1)

$x (x^2 + 3y^2 - 1) = 0$ ...(2)

$y = 0$ satisfies (1). That solution leaves you with $x = 0$ as solution or else $(x^2 - 1) = 0$, i.e. $x = \pm1$

Without further ado by the symmetry between the equations you can now say that also $x = 0, y = \pm1$ must be a solution.

For the totally non-zero solutions for every x/y that satisfies (3), the same value of y/x satisfies (4).

$(y^2 + 3x^2 - 1) = 0$ ...(3)

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

So $\frac{x}{y} = \frac{y}{x}$

will be in the non-zero solutions. So either $$x = y$$ or $$x = -y$$. So you can substitute those in one of (3, 4) and have something simple.

This does not save all that much work but gives more insight and shows results you get are not an accident. Mathematicians are always looking for symmetries in equations etc., I think they might tell us they do nothing else.

Edit: solving the first way will give you a guide through the cleverer second way. Mathematicians often do that too!

Do not count I have got it completely right, even complete and even right!

Last edited: Apr 12, 2010
10. Apr 12, 2010

### Koldstream

Thankyou! But i still need to study it to convince myself of it totally.

11. Apr 12, 2010

### epenguin

Yes that is exactly what you need to do.

So did I again, there is possibly not much difference between different methods. This time I get four non-zero solutions quite easily.

You mentioned this was part of another problem and probably that could itself be simplified using symmetry.