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v4xN0s
 3 years ago
write a differential equation in dy/dt=ay+b form with all solutions approaching y=3 as t> infinity
v4xN0s
 3 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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vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Actually, we can reason it out another way.

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know what dy/dt *means*?

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Okay then, so if limt>infy should give y(t) = 3, then, at infinity, what should the slope of y(t) be?

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1If, at infinity, our function needs to keep getting closer to y=3, then what value should the SLOPE of the function be approaching?

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1@v4xN0s Yes, 0. What's another name for slope?

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, or, more relavant to this problem, dy/dt. Now, as you said, we want dy/dt = 0 when y = 3, right?

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

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

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1@zzr0ck3r yes thats what the question says

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0what if all the solutions diverged from y=2

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1"diverged from y = 2" > what?

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0as t approaches infinity

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1@v4xN0s we're not done with the previous one yet

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0oh alright i thought that was the write answer QQ

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1But it depends on the initial conditions, too.

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see well i just needed one diff eq for the problem

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1See, 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
 3 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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Yes. If\[y_0=3\], then your solution is a line, starting at y=3 and staying that way (dy/dt = 0).

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0wait no slope still decreases

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah, those are the three possible solution curves. Why do you think the slope decreases for the initial cond less than 3 one?

v4xN0s
 3 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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, its *decreasing* but *positive*. That's the whole deal  regardless where it is, it gets closer to 0 as it goes to infinity.

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

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1So, 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)?

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

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

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0its decreasing and negative

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1No. 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.

v4xN0s
 3 years ago
Best ResponseYou've already chosen the best response.0sorry but i have no idea

vf321
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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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