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

- anonymous

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

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- anonymous

no im dumb

- anonymous

Actually, we can reason it out another way.

- anonymous

Do you know what dy/dt *means*?

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## More answers

- zzr0ck3r

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

- anonymous

ya i do

- anonymous

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

- anonymous

3?

- anonymous

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

- anonymous

infinity

- anonymous

wait no 0

- zzr0ck3r

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

- anonymous

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

- anonymous

derivative

- anonymous

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

- anonymous

yep

- anonymous

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

- anonymous

so it would be y'=3-y?

- anonymous

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

- zzr0ck3r

no we want y = 3 when t = infinity?

- anonymous

@zzr0ck3r yes thats what the question says

- anonymous

what if all the solutions diverged from y=2

- anonymous

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

- anonymous

as t approaches infinity

- anonymous

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

- anonymous

oh alright i thought that was the write answer QQ

- anonymous

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

- anonymous

But it depends on the initial conditions, too.

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

wait no slope still decreases

- anonymous

Um no.....

- anonymous

|dw:1345523728600:dw|

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)?

- anonymous

it becomes negative?

- anonymous

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

- anonymous

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

- anonymous

its decreasing and negative

- anonymous

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.

- anonymous

sorry but i have no idea

- anonymous

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?

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