Help with easy Differential Equation

[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?

 

[TD]

Looks good

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

 

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

 

[TD]

You leave it in. But honestly, me suggestion was meant 'in general', I now realize 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

 

[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.

 

[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.

 

[TD]

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

That's great

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

I think I already did

 

[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:
$$y \prime = F(x,y),$$
$$y(x_{0})=y(y_{0})$$

 

[JoshHolloway]

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

 

[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).

 

[JoshHolloway]

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

yes, your right.

 

[JoshHolloway]

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

 

[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?

 

[TD]

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

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.

 

[TD]

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

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

 

[JoshHolloway]

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

 

[JoshHolloway]

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

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)?

 

[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).

 

[JoshHolloway]

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.

My problem does not have an initial condition.

 

[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?

 

[TD]

I believe so, but without the point (x0,y0) given, you cannot check if this region contains the point of course.

 

[JoshHolloway]

I get it. Thanks.

 

[JoshHolloway]

Is the answer ALL X SUCH THAT X DOES NOT EQUAL ZERO?

How do I know when the solution is not differentiable if the C is in the argument and I am not given an initial condition?

 

[TD]

I believe x is the argument in your solution y(x).

 

[JoshHolloway]

I believe x is the argument in your solution y(x).

What do you mean? Aren't I supposed to find the values of x for which the solution is differentiable? Wouldn't that be all values of x except when x is zero (or when it is coterminal with zero)?

 

[TD]

I'm sorry, I misread your last post as if you thought 'C is the argument', I missed the "in". You're right, it'll depend on C.

You have something of the form y = tan(f(x,c)). Now, tan(a) is nicely continuous and differentiable unless a is an integer multiple of pi (you have a vertical asymptote there). So you need to check where this f(x,c) = k*pi.

Normally, the 'C' vanishes thanks to your condition of the initial value problem, f(x0) = y0. Apparently, that was given here (which I find a bit odd, since it's a part of the conditions you had to check in the existence theorem).

 

[JoshHolloway]

Cool, cool. I think you have pointed me in the direction I needed. Thanks a lot buddy.

 

[TD]

You're welcome