# Differential Equations Problem

1. Sep 18, 2005

### amcavoy

I have the following problem:

$$y'=\frac{3x^{2}}{3y^{2}-4};\quad y(1)=0$$

I solved this, and came up with $y^{3}-4y-x^{3}=-1$

My problem lies in finding the interval of definition. From the original DE, it is clear that $|y|\neq \frac{2\sqrt{3}}{3}$. Then, after plugging that in for y and solving for what x can't be, I come up with:

$$\left|x^{3}-1\right|\neq \frac{16\sqrt{3}}{9}$$

...however, my reasoning doesn't seem to work here. Apparently, the answer is:

$$\left|x^{3}-1\right|<\frac{16\sqrt{3}}{3}$$

Can someone help me out here?

Thanks a lot.

Last edited: Sep 18, 2005
2. Sep 18, 2005

### arildno

Well, the y-value can't be allowed to slide past the point of singularity when starting on 1, can it?

3. Sep 18, 2005

### amcavoy

Alright, well that explains why it is less-than, but why is the denominator 3 instead of 9? Each time I work it out, I come up with 9 in the denominator.

4. Sep 18, 2005

### Hurkyl

Staff Emeritus

5. Sep 18, 2005

### amcavoy

$$\left|x^{3}-1\right|=y\left(y^{2}-4\right)$$

I then plug in $y=\frac{2\sqrt{3}}{3}$ (which is one of the values y cannot be) to see what value(s) of x would give this:

$$\left|x^{3}-1\right|\neq\frac{2\sqrt{3}}{3}\left(\frac{4}{3}-4\right)$$

..and upon solving for this particular value, I come up with $\left|x^{3}-1\right|\neq\frac{16\sqrt{3}}{9}$, which is incorrect. Why is this?

Thanks again.

6. Sep 18, 2005

### saltydog

$$\frac{16}{3\sqrt{3}}$$

is correct. Can you generate an implicit plot of the solution? I've attached one. Do you understand why it looks like that and why the ENTIRE plot satisfies the differential equation even at the points of vertical tangency? I included the two vertical lines to show where the values of x are that you calculated.

#### Attached Files:

• ###### odeplot1.JPG
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7. Sep 18, 2005

### Hurkyl

Staff Emeritus
Well, I presume you meant to say:

$$\left|x^{3}-1\right|=\left|y\left(y^{2}-4\right)\right|$$

Anyways, here's a problem:

You know that y cannot be equal to ±2√3/3.

However, that does not mean that y(y²-4) cannot be equal to

(2√3/3)((2√3/3)²-4)

Last edited: Sep 18, 2005
8. Sep 18, 2005

### amcavoy

Ok that makes sense. Thank you everyone for your help. By the way, which program did you use to generate that plot? I use Linux and can't find anything that is relatively inexpensive or free for graphing.

Thanks again.

9. Sep 18, 2005

### saltydog

I used Mathematica but I'm sure others would do. I tell you what, suppose you just absolutely had to write from scratch in C++, a program to plot it. You could do that right? Just for each value of x in the domain, solve the cubic polynomial for y, get the (real) roots, put them all together, and there you go.

10. Sep 18, 2005

### Hurkyl

Staff Emeritus
Bleh, if I had remembered it was a point of tangency, I could have found the other value of y much more easily when x³-1 = 16√3/9. (Oh well, at least I only had to solve a quadratic!)

11. Sep 18, 2005

### saltydog

Just for the record: It's easier to solve for x in terms of y (at least to get the plot). Just for each value of y in some range, compute:

$$x=\sqrt[3]{1+y^3-4y}$$

Ok, I'm done.

12. Sep 18, 2005

### amcavoy

Thank you so much for all the help! I appreciate that graph. Now I can see why it wouldn't be a function if the range contained anything else.