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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
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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vf321 Group TitleBest ResponseYou've already chosen the best response.1
Actually, we can reason it out another way.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Do you know what dy/dt *means*?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
y = (3n+1)/(n+1) dy/dt = 2/(x+1)^2
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
what is the limit of y = (3n+1)/(n+1) at infinity?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
@v4xN0s Yes, 0. What's another name for slope?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Okay then. In ay+b, if I make y = 3, then what can a and b be?
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
so it would be y'=3y?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
That's one solution, b = 3 and a = 1.
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.0
no we want y = 3 when t = infinity?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
@zzr0ck3r yes thats what the question says
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
what if all the solutions diverged from y=2
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
"diverged from y = 2" > what?
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
as t approaches infinity
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
@v4xN0s we're not done with the previous one yet
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
oh alright i thought that was the write answer QQ
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
But it depends on the initial conditions, too.
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
oh i see well i just needed one diff eq for the problem
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Okay, your example of dy/dt=3y was a bit of a lucky guess.
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
i see and if its less than 3 then its positive and increases as it gets closer to 3
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
wait no slope still decreases
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
dw:1345523728600:dw
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
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

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
ok got that one, but what if solutions diverge from y=2
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
So, back to where I was. You were lucky when you picked dy/dt as 3y. What would have happened if you've done y3 (note, this still has slope of 0 when y =3)?
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
it becomes negative?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Well think about it. If our \[y_0>3\]What will happen?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
We're above the line we want to tend to in end behavior, and what's our slope?
 one year ago

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
its decreasing and negative
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
No. if y>3, and dy/dt is y3, our slope is positive. As y increases, y3 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

v4xN0s Group TitleBest ResponseYou've already chosen the best response.0
sorry but i have no idea
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Well, if y > 3, do you see why y3 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 y3 also gets bigger). Do you not see this makes you diverge?
 one year ago
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