# Help with easy Differential Equation

1. May 31, 2006

### JoshHolloway

Here is the equation:
$$y\prime=1+x+y^2+xy^2$$

I have gotten it to this point:
$$\tan^{-1}(y) = x + \frac{x^{2}}{2} + C$$

Now I am not sure where to go from here. Am I supposed to take the tangent of both sides like this:
$$tan(tan^{-1}(y)) = tan(x + \frac{x^{2}}{2} + C)$$

$$y = tan(x + \frac{x^{2}}{2} + C)$$

Is that the solution?

2. May 31, 2006

### TD

Looks good

If you're unsure, you can always check your answer by plugging it back in the original DE.

3. May 31, 2006

### JoshHolloway

When I am plugging the general solution back into the original eqaution to test it do I include the integration constant? Or do I leave it out? What I mean is do I plug this in:
$$y = tan(x + \frac{x^{2}}{2} + C)$$

Or do I plug this in:
$$y = tan(x + \frac{x^{2}}{2})$$

4. May 31, 2006

### TD

You leave it in. But honestly, me suggestion was meant 'in general', I now realise that it might be an annoying task to put this back in and simplify to see if it's correct. Sometimes this is a fairly easy method to check your answer. Anyway: your solution is correct

5. May 31, 2006

### JoshHolloway

Ha Ha. I really went and took your advice and worked the damned thing all the way to the end! I ended up with $$1+x=1+x$$
Thanks for the help.

Hey if you don't mind could you check out the thread I just started in the homework section under the precalculus maths. Thanks.

6. May 31, 2006

### JoshHolloway

Another question:
State the regions of the xy-plane in which the conditions of the fundamental existence and uniqueness theorem are satisfied.

I have no idea how to determine this. Could you possibly point me in the correct dirrection. In just a second I will post the theorem if you don't know it.

7. May 31, 2006

### TD

That's great

8. May 31, 2006

### JoshHolloway

Here is the theorem:
Let F and $$\frac{\partial F}{\partial y}$$ be continuous functions in some rectangle $$\alpha < x <\beta ,\ \gamma < y < \delta$$ containing the point $$(x_{0},y_{0})$$. Then, in some interval $$(x-h, x+h) \subset (\alpha, \beta)$$, there is a unique solution to the intial value problem:
$$y \prime = F(x,y),$$
$$y(x_{0})=y(y_{0})$$

Last edited: May 31, 2006
9. May 31, 2006

### JoshHolloway

I really don't understand this theorum. Is there any chance you could explain it to me?

Last edited: May 31, 2006
10. May 31, 2006

### TD

I assume delta and gamma are the boundaries for y, and not for x again?

The theorem states that if the required criteria are met, i.e. if F is differentiable and F' continuous (is this necessary?) on a certain area which contains a point (x0,y0), then the differential equation y' = F(x,y) with initial value y(x0) = y0 has a unique solution in the neighbourhood of (x0,y0).

11. May 31, 2006

### JoshHolloway

12. May 31, 2006

### JoshHolloway

What do you mean by "in the neighbourhood"? Do you mean just close to that point?

13. May 31, 2006

### JoshHolloway

So here is the question:
State the regions of the xy-plane in which the conditions of the fundamental existence and uniqueness theorem are satisfied.

Is that basically asking on what interval is the solution differentianle and continious?

14. May 31, 2006

### TD

Yes, that's a typical expression in mathematics. It's always an open set, in general it's an open ball of some dimension with a radius r > 0. In the neighbourhood of a point x0 would mean an interval (x0-e,x0+e) where e > 0.

It means that the solution exists at least at the point x0 and 'very close to it'. Often the solution will be valid on a larger domain.

15. May 31, 2006

### TD

Yes, and the point (x0,y0) has to be in the interior of this domain.

16. May 31, 2006

### JoshHolloway

I understand that about 60 percent of the way. But I don't really know what an open set is.

Last edited: Jun 1, 2006
17. May 31, 2006

### JoshHolloway

Am I actually supposed to find the point (x0,y0). I don't really understand that part. I understand that I am supposed to state the domain in which the solution is differentiable and continious. But what does that have to do with some point (x0,y0)?

Last edited: May 31, 2006
18. May 31, 2006

### TD

For an intial value problem, that point (x0,y0) should be given. This also allows you to determine the constant C, hence yielding a unique solution. Your solution isn't unique yet, you have 'infinite solutions', because of that indetermined C.

The 'open' means that for the interval (x0-e,x0+e), the end points aren't in the interval, as opposed to a closed interval [x0-e,x0+e]. The open ball I mentioned is for such neighbourhoods in general, in higher dimensions for example. There you have the point x0 as center and a radius r (or e, as you wish).

19. May 31, 2006

### JoshHolloway

My problem does not have an initial condition.

20. May 31, 2006

### JoshHolloway

It says state the region in which the theorem is satisfied. So I should just state the interval in which the solution is differentiable and continious?