# Differential Equations Problem

• amcavoy
In summary, the conversation discusses a problem involving a differential equation and finding the interval of definition. The solution for the problem is y^3-4y-x^3=-1, but there is confusion about finding the correct interval. The conversation also includes a plot of the solution and mentions using Mathematica for graphing.
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:
Well, the y-value can't be allowed to slide past the point of singularity when starting on 1, can it?

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.

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

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.

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

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.

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

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.

#### Attachments

• odeplot1.JPG
3.5 KB · Views: 347
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:
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.

apmcavoy said:
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.

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.

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!)

saltydog said:
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.

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.

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.

## 1. What is a differential equation?

A differential equation is a mathematical equation that involves an unknown function and its derivatives. It describes the relationship between the rate of change of a variable and the values of the variable itself.

## 2. Why are differential equations important?

Differential equations are important because they are used to model real-life situations in fields such as physics, engineering, and economics. They allow us to predict how a system will behave over time and make informed decisions based on those predictions.

## 3. What are the types of differential equations?

There are several types of differential equations, including ordinary differential equations (ODEs), partial differential equations (PDEs), and stochastic differential equations (SDEs). ODEs involve only one independent variable, PDEs involve multiple independent variables, and SDEs incorporate randomness into the equation.

## 4. How do you solve a differential equation?

The method for solving a differential equation depends on its type and complexity. Some methods include separation of variables, substitution, and using specific formulas for certain types of equations. In some cases, a differential equation cannot be solved analytically and numerical methods must be used.

## 5. What are some applications of differential equations?

Differential equations have a wide range of applications in various fields. Some common applications include modeling population growth, predicting weather patterns, designing electrical circuits, and understanding the behavior of chemical reactions. They are also used in economics, biology, and many other areas of science and engineering.

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