Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

no im dumb
Actually, we can reason it out another way.
Do you know what dy/dt *means*?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

y = (3n+1)/(n+1) dy/dt = 2/(x+1)^2
ya i do
Okay then, so if limt->infy should give y(t) = 3, then, at infinity, what should the slope of y(t) be?
3?
If, at infinity, our function needs to keep getting closer to y=3, then what value should the SLOPE of the function be approaching?
infinity
wait no 0
what is the limit of y = (3n+1)/(n+1) at infinity?
@v4xN0s Yes, 0. What's another name for slope?
derivative
Yes, or, more relavant to this problem, dy/dt. Now, as you said, we want dy/dt = 0 when y = 3, right?
yep
Okay then. In ay+b, if I make y = 3, then what can a and b be?
so it would be y'=3-y?
That's one solution, b = 3 and a = -1.
no we want y = 3 when t = infinity?
@zzr0ck3r yes thats what the question says
what if all the solutions diverged from y=2
"diverged from y = 2" --> what?
as t approaches infinity
@v4xN0s we're not done with the previous one yet
oh alright i thought that was the write answer QQ
QQ? Anyway, that's just ONE anwer. In fact, any a and b that makes 3a+b = 0 works.
But it depends on the initial conditions, too.
oh i see well i just needed one diff eq for the problem
Okay, your example of dy/dt=3-y was a bit of a lucky guess.
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.
i see and if its less than 3 then its positive and increases as it gets closer to 3
Yes. If\[y_0=3\], then your solution is a line, starting at y=3 and staying that way (dy/dt = 0).
wait no slope still decreases
Um no.....
|dw:1345523728600:dw|
Yeah, those are the three possible solution curves. Why do you think the slope decreases for the initial cond less than 3 one?
the bottom curve's slope is appreaching 0 isnt it, so the slope would be decreasing
Yes, its *decreasing* but *positive*. That's the whole deal - regardless where it is, it gets closer to 0 as it goes to infinity.
ok got that one, but what if solutions diverge from y=2
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)?
it becomes negative?
Well think about it. If our \[y_0>3\]What will happen?
We're above the line we want to tend to in end behavior, and what's our slope?
its decreasing and negative
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.
sorry but i have no idea
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?

Not the answer you are looking for?

Search for more explanations.

Ask your own question