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no im dumb

Actually, we can reason it out another way.

Do you know what dy/dt *means*?

y = (3n+1)/(n+1)
dy/dt = 2/(x+1)^2

ya i do

3?

infinity

wait no 0

what is the limit of y = (3n+1)/(n+1) at infinity?

derivative

yep

Okay then. In ay+b, if I make y = 3, then what can a and b be?

so it would be y'=3-y?

That's one solution, b = 3 and a = -1.

no we want y = 3 when t = infinity?

what if all the solutions diverged from y=2

"diverged from y = 2" --> what?

as t approaches infinity

oh alright i thought that was the write answer QQ

QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.

But it depends on the initial conditions, too.

oh i see well i just needed one diff eq for the problem

Okay, your example of dy/dt=3-y was a bit of a lucky guess.

i see and if its less than 3 then its positive and increases as it gets closer to 3

Yes. If\[y_0=3\], then your solution is a line, starting at y=3 and staying that way (dy/dt = 0).

wait no slope still decreases

Um no.....

|dw:1345523728600:dw|

the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing

ok got that one, but what if solutions diverge from y=2

it becomes negative?

Well think about it. If our \[y_0>3\]What will happen?

We're above the line we want to tend to in end behavior, and what's our slope?

its decreasing and negative

sorry but i have no idea