Diff EQ Direction field asymptotes

[ElijahRockers]

Homework Statement

Draw the direction field for the differential equation y'=1-y/x

Homework Equations

The Attempt at a Solution

Ok well, drawing the direction field is not an issue because I have a grapher, and I get the basic of how to draw simple direction fields. So to start, I know that whenever y=x, the slope will be zero, so every point on the line y=x has slope zero. I can test points around that to get a general idea of what's happening. Also I know when x approaches zero, the slope approached -/+ infinity depending on the sign of y.

So far so good...

But the other asymptote besides x=0 also seems to be a line y=x/2. How do I conclude that analytically from y'=1-y/x ? I'm sure it's something simple...

Thanks.

 

[HallsofIvy]

If y= ax is an asymptote then points on that line must be mapped to other points of that line. y'= 1- y/x is the same as y'= (x- y)/x and if y= ax, that becomes a= y'= (x-ax)/x= 1- a. y= ax will be an asyptote if a= 1- a or a= 1/2. That is, the asymptote is y= (1/2)x as you say.

 

[ElijahRockers]

yikes, ok i see how you worked it out, i don't see any voodoo but it certainly blew my mind when you went from y=ax to y'=a, and subsequently to a=1-a...

ok, so I'm not quite clear on what you mean by If y= ax is an asymptote then points on that line must be mapped to other points of that line.

he probably won't make us do very complicated ones, but i am interested on how i could translate this technique (if possible) into finding asymptotes described by polynomials, for instance?

thanks for all your help on the problem, if you know of any resources that discuss drawing direction fields in more detail i would be interested... every example i see is usually a simple one with only horizontal asymptotes

thanks again