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v4xN0s Group Title

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

  • one year ago
  • one year ago

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  1. v4xN0s Group Title
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    no im dumb

    • one year ago
  2. vf321 Group Title
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    Actually, we can reason it out another way.

    • one year ago
  3. vf321 Group Title
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    Do you know what dy/dt *means*?

    • one year ago
  4. zzr0ck3r Group Title
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    y = (3n+1)/(n+1) dy/dt = 2/(x+1)^2

    • one year ago
  5. v4xN0s Group Title
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    ya i do

    • one year ago
  6. vf321 Group Title
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    Okay then, so if limt->infy should give y(t) = 3, then, at infinity, what should the slope of y(t) be?

    • one year ago
  7. v4xN0s Group Title
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    3?

    • one year ago
  8. vf321 Group Title
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    If, at infinity, our function needs to keep getting closer to y=3, then what value should the SLOPE of the function be approaching?

    • one year ago
  9. v4xN0s Group Title
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    infinity

    • one year ago
  10. v4xN0s Group Title
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    wait no 0

    • one year ago
  11. zzr0ck3r Group Title
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    what is the limit of y = (3n+1)/(n+1) at infinity?

    • one year ago
  12. vf321 Group Title
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    @v4xN0s Yes, 0. What's another name for slope?

    • one year ago
  13. v4xN0s Group Title
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    derivative

    • one year ago
  14. vf321 Group Title
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    Yes, or, more relavant to this problem, dy/dt. Now, as you said, we want dy/dt = 0 when y = 3, right?

    • one year ago
  15. v4xN0s Group Title
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    yep

    • one year ago
  16. vf321 Group Title
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    Okay then. In ay+b, if I make y = 3, then what can a and b be?

    • one year ago
  17. v4xN0s Group Title
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    so it would be y'=3-y?

    • one year ago
  18. vf321 Group Title
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    That's one solution, b = 3 and a = -1.

    • one year ago
  19. zzr0ck3r Group Title
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    no we want y = 3 when t = infinity?

    • one year ago
  20. vf321 Group Title
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    @zzr0ck3r yes thats what the question says

    • one year ago
  21. v4xN0s Group Title
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    what if all the solutions diverged from y=2

    • one year ago
  22. vf321 Group Title
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    "diverged from y = 2" --> what?

    • one year ago
  23. v4xN0s Group Title
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    as t approaches infinity

    • one year ago
  24. vf321 Group Title
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    @v4xN0s we're not done with the previous one yet

    • one year ago
  25. v4xN0s Group Title
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    oh alright i thought that was the write answer QQ

    • one year ago
  26. vf321 Group Title
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    QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.

    • one year ago
  27. vf321 Group Title
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    But it depends on the initial conditions, too.

    • one year ago
  28. v4xN0s Group Title
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    oh i see well i just needed one diff eq for the problem

    • one year ago
  29. vf321 Group Title
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    Okay, your example of dy/dt=3-y was a bit of a lucky guess.

    • one year ago
  30. vf321 Group Title
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    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.

    • one year ago
  31. v4xN0s Group Title
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    i see and if its less than 3 then its positive and increases as it gets closer to 3

    • one year ago
  32. vf321 Group Title
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    Yes. If\[y_0=3\], then your solution is a line, starting at y=3 and staying that way (dy/dt = 0).

    • one year ago
  33. v4xN0s Group Title
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    wait no slope still decreases

    • one year ago
  34. vf321 Group Title
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    Um no.....

    • one year ago
  35. v4xN0s Group Title
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    |dw:1345523728600:dw|

    • one year ago
  36. vf321 Group Title
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    Yeah, those are the three possible solution curves. Why do you think the slope decreases for the initial cond less than 3 one?

    • one year ago
  37. v4xN0s Group Title
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    the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing

    • one year ago
  38. vf321 Group Title
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    Yes, its *decreasing* but *positive*. That's the whole deal - regardless where it is, it gets closer to 0 as it goes to infinity.

    • one year ago
  39. v4xN0s Group Title
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    ok got that one, but what if solutions diverge from y=2

    • one year ago
  40. vf321 Group Title
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    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)?

    • one year ago
  41. v4xN0s Group Title
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    it becomes negative?

    • one year ago
  42. vf321 Group Title
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    Well think about it. If our \[y_0>3\]What will happen?

    • one year ago
  43. vf321 Group Title
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    We're above the line we want to tend to in end behavior, and what's our slope?

    • one year ago
  44. v4xN0s Group Title
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    its decreasing and negative

    • one year ago
  45. vf321 Group Title
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    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.

    • one year ago
  46. v4xN0s Group Title
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    sorry but i have no idea

    • one year ago
  47. vf321 Group Title
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    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?

    • one year ago
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