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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, we can reason it out another way.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you know what dy/dt *means*?

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0y = (3n+1)/(n+1) dy/dt = 2/(x+1)^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay then, so if limt>infy should give y(t) = 3, then, at infinity, what should the slope of y(t) be?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If, at infinity, our function needs to keep getting closer to y=3, then what value should the SLOPE of the function be approaching?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@v4xN0s Yes, 0. What's another name for slope?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, or, more relavant to this problem, dy/dt. Now, as you said, we want dy/dt = 0 when y = 3, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay then. In ay+b, if I make y = 3, then what can a and b be?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so it would be y'=3y?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's one solution, b = 3 and a = 1.

zzr0ck3r
 4 years ago
Best ResponseYou've already chosen the best response.0no we want y = 3 when t = infinity?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zzr0ck3r yes thats what the question says

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what if all the solutions diverged from y=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"diverged from y = 2" > what?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as t approaches infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@v4xN0s we're not done with the previous one yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh alright i thought that was the write answer QQ

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But it depends on the initial conditions, too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i see well i just needed one diff eq for the problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, your example of dy/dt=3y was a bit of a lucky guess.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0See, 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
 4 years ago
Best ResponseYou've already chosen the best response.0i see and if its less than 3 then its positive and increases as it gets closer to 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. If\[y_0=3\], then your solution is a line, starting at y=3 and staying that way (dy/dt = 0).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait no slope still decreases

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1345523728600:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, those are the three possible solution curves. Why do you think the slope decreases for the initial cond less than 3 one?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, its *decreasing* but *positive*. That's the whole deal  regardless where it is, it gets closer to 0 as it goes to infinity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok got that one, but what if solutions diverge from y=2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, 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)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well think about it. If our \[y_0>3\]What will happen?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We're above the line we want to tend to in end behavior, and what's our slope?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its decreasing and negative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry but i have no idea

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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?
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