Differential equation asymptotes

lpbug
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Hi guys, I just have a very broad and general question.
Today in math class I was asked to solve the horizontal asymptote of a differential equation, and this had me stumped.

Later on, the teacher gave me the following two examples:

dy/dx=x-2
and
dy/dx=y-2

The solution to the first equation for a horizontal asymptote is DNE
The solution to the second is 2

Now, I'm just wondering... How would one know that this is true without solving for the original equation? I mean, this doesn't seem intuitive at all to me. Why is it that when X is what makes the differential equation 0 there is no asymptote and when Y makes the equation 0 there is?

Thanks for all the help.
 
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If you have a horizontal asymptote then as x approaches infinity, dy/dx has to approach zero, right?
 
but isn't it not enough information just to assume that whenever dy/dx=0 when the y value is making the dy/dx 0? I mean, I see where you're comming from with as x approaches negative infinity or infinity dy/dx has to approach 0 but I don't understand how you can just tell from the equation that a dy/dx involving a y will DEFINITELY have a horizontal asymptote. Is it because all forms of differential equation involving y will have a solution like e^something?
 
Also, the reason for dy/dx being zero if y is a certain number implying a horizontal asymptote is simple: If when y is a certain number, then dy/dx is zero, then the graph is going to be flat at that point. This means that y won't change as x changes, but since y doesn't change, then dy/dx is going to stay zero. Hence, horizontal asymptote.
 
aha! I think i got it, so if the slope of dy/dx is 0, either the change in y (dy) must equal 0 OR the change in x (dx) must be infinity? and if the y isn't changing, then the slope will not change after it theoretically reaches 0, because dy/dx is dependent on y itself?
 
lpbug said:
but isn't it not enough information just to assume that whenever dy/dx=0 when the y value is making the dy/dx 0? I mean, I see where you're comming from with as x approaches negative infinity or infinity dy/dx has to approach 0 but I don't understand how you can just tell from the equation that a dy/dx involving a y will DEFINITELY have a horizontal asymptote. Is it because all forms of differential equation involving y will have a solution like e^something?

Well, you can definitely say that the first equation doesn't have an asymptote. In the second equation there is at least a possible y value for a horizontal asymptote. One way to look at it is if y>2 then the function y is increasing. If y<2 it's decreasing. Imagine what must happen as x->-infinity.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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