v4xN0s
write a differential equation in dy/dt=ay+b form with all solutions approaching y=3 as t-> infinity
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v4xN0s
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no im dumb
vf321
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Actually, we can reason it out another way.
vf321
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Do you know what dy/dt *means*?
zzr0ck3r
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y = (3n+1)/(n+1)
dy/dt = 2/(x+1)^2
v4xN0s
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ya i do
vf321
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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?
v4xN0s
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3?
vf321
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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?
v4xN0s
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infinity
v4xN0s
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wait no 0
zzr0ck3r
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what is the limit of y = (3n+1)/(n+1) at infinity?
vf321
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@v4xN0s Yes, 0. What's another name for slope?
v4xN0s
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derivative
vf321
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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?
v4xN0s
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yep
vf321
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Okay then. In ay+b, if I make y = 3, then what can a and b be?
v4xN0s
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so it would be y'=3-y?
vf321
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That's one solution, b = 3 and a = -1.
zzr0ck3r
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no we want y = 3 when t = infinity?
vf321
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@zzr0ck3r yes thats what the question says
v4xN0s
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what if all the solutions diverged from y=2
vf321
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"diverged from y = 2" --> what?
v4xN0s
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as t approaches infinity
vf321
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@v4xN0s we're not done with the previous one yet
v4xN0s
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oh alright i thought that was the write answer QQ
vf321
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QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.
vf321
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But it depends on the initial conditions, too.
v4xN0s
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oh i see well i just needed one diff eq for the problem
vf321
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Okay, your example of dy/dt=3-y was a bit of a lucky guess.
vf321
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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.
v4xN0s
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i see and if its less than 3 then its positive and increases as it gets closer to 3
vf321
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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).
v4xN0s
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wait no slope still decreases
vf321
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Um no.....
v4xN0s
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|dw:1345523728600:dw|
vf321
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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?
v4xN0s
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the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing
vf321
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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.
v4xN0s
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ok got that one, but what if solutions diverge from y=2
vf321
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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)?
v4xN0s
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it becomes negative?
vf321
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Well think about it. If our \[y_0>3\]What will happen?
vf321
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We're above the line we want to tend to in end behavior, and what's our slope?
v4xN0s
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its decreasing and negative
vf321
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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.
v4xN0s
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sorry but i have no idea
vf321
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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?