## v4xN0s 3 years ago write a differential equation in dy/dt=ay+b form with all solutions approaching y=3 as t-> infinity

1. v4xN0s

no im dumb

2. vf321

Actually, we can reason it out another way.

3. vf321

Do you know what dy/dt *means*?

4. zzr0ck3r

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

5. v4xN0s

ya i do

6. vf321

Okay then, so if limt->infy should give y(t) = 3, then, at infinity, what should the slope of y(t) be?

7. v4xN0s

3?

8. vf321

If, at infinity, our function needs to keep getting closer to y=3, then what value should the SLOPE of the function be approaching?

9. v4xN0s

infinity

10. v4xN0s

wait no 0

11. zzr0ck3r

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

12. vf321

@v4xN0s Yes, 0. What's another name for slope?

13. v4xN0s

derivative

14. vf321

Yes, or, more relavant to this problem, dy/dt. Now, as you said, we want dy/dt = 0 when y = 3, right?

15. v4xN0s

yep

16. vf321

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

17. v4xN0s

so it would be y'=3-y?

18. vf321

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

19. zzr0ck3r

no we want y = 3 when t = infinity?

20. vf321

@zzr0ck3r yes thats what the question says

21. v4xN0s

what if all the solutions diverged from y=2

22. vf321

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

23. v4xN0s

as t approaches infinity

24. vf321

@v4xN0s we're not done with the previous one yet

25. v4xN0s

oh alright i thought that was the write answer QQ

26. vf321

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

27. vf321

But it depends on the initial conditions, too.

28. v4xN0s

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

29. vf321

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

30. vf321

See, if your initial condition is\[y_0>3\]then your slope is negative, and y decreases until it 'reaches' 3, at which point it 'stays' constant.

31. v4xN0s

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

32. vf321

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

33. v4xN0s

wait no slope still decreases

34. vf321

Um no.....

35. v4xN0s

|dw:1345523728600:dw|

36. vf321

Yeah, those are the three possible solution curves. Why do you think the slope decreases for the initial cond less than 3 one?

37. v4xN0s

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

38. vf321

Yes, its *decreasing* but *positive*. That's the whole deal - regardless where it is, it gets closer to 0 as it goes to infinity.

39. v4xN0s

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

40. vf321

So, back to where I was. You were lucky when you picked dy/dt as 3-y. What would have happened if you've done y-3 (note, this still has slope of 0 when y =3)?

41. v4xN0s

it becomes negative?

42. vf321

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

43. vf321

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

44. v4xN0s

its decreasing and negative

45. vf321

No. if y>3, and dy/dt is y-3, our slope is positive. As y increases, y-3 increases. Thus dy/dt is increasing and positive. What do you think this will do to the function? Then try y_0=3 and y_0<3.

46. v4xN0s

sorry but i have no idea

47. vf321

Well, if y > 3, do you see why y-3 is > 0? Then dy/dt is greater than 0. Thus our slope is positive. Our particle or whatever increases in y in that differential amount of time, resulting in an even larger slope (since y-3 also gets bigger). Do you not see this makes you diverge?