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anonymous
 5 years ago
How would I use the Bernoulli Equation to solve t^2dy/dt+2tyy^3=0, t>0?
anonymous
 5 years ago
How would I use the Bernoulli Equation to solve t^2dy/dt+2tyy^3=0, t>0?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The basic idea is to move y^3 to the other side. Then you might recognize Left Hand Side is just the derivative of y * t^2 with respect to t. Then you can integrate both sides with respect to t. On the Left Hand Side, you get y * t^2. On the Right Hand Side, you get t*y^3 + C. Then the solution will be curves y(t) that satisfy y*t^2 – t*y^3 = C for some arbitrary C. duongni asked that about 2 hrs ago. I responded to it about half an hour ago. If you want the details, you can scroll down there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you very, very much, verifry. You're answer to duongi's curiously identical question was immensely helpful.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm having a really hard time with the check on this one. would you be willing to walk me through it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You mean checking the solution to see if it solves the original equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I keep ending up with a 3ty^2dy/dt that doesn't fit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you be willing to walk me through that part as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You know what, I think what I wrote isn't the solution anymore. I'm sorry. I thought it checked out, but you're right when you say it doesn't fit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The problem is that when we integrated y^3, we shouldn't just get ty^3 + C, since y isn't a constant, it's actually a function of t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How should it look then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me work it out and check it this time before typing anything.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool beans. Thank you again for sticking with me through all of this. I really appreciate the help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying to check it right now, but the check is not pretty. The basic method of solving it is better explained here: http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx The idea is to convert it down to an equation in which we can apply the method of integrating factors without having a function of y on the right hand side.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The reason the method of integrating factors didn't work above in my first explanation was because the Right Hand Side had a function of y, so when we integrated it, we can't solve the integral explicitly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The reason the method of integrating factors didn't work above in my first explanation was because the Right Hand Side had a function of y, so when we integrated it, we can't solve the integral explicitly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if we divide both sides by y^3 first, we can make the Right Side not a function of y anymore. Then we try a substitution to convert the differential equation to one in which it's easier to see how we can use the method of integrating factors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you be willing to walk me through that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Starting from t^2 * y' + 2ty = y^3, we want to divide by y^3 to get the RHS to not be a function of y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we get t^2 * (y'/y^3) + 2t / y^2 = 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now we want to make a change of variables so that it's easier to do the method of integrating factors. Maybe we'd like our new variable to be v, and we want v so that v' is some constant multiple of y'/y^3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Normally, after solving lots of these problems and generalizing the method, you'd just let v = y^(1n) because you've seen the pattern. But if you haven't seen the pattern, then we might as well just let v' be equal to y'/y^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why can't we use v=y^(1n)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We can. We probably should to make it simplify. I was just thinking that if this was the first time you've seen this, letting v = y^(1n) might seem to pop out of nowhere.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I sort of read through the website explanation, but I just don't pick stuff up very well that way. We could use v=y^(1n) though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah let's let v = y^(1n). Here our n = 3, so let's let v = y^(2). So then dv = (2) y^(3) dy. So dv/dt = (2) y^(3) dy/dt.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you aren't sure of the answer you should visit aplha math help its on google, and shows you the exact answer!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow, alpha math help is pretty cool. I didn't know about that either.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://blog.wolframalpha.com/2009/12/01/stepbystepmath/ But I don't know if it can do differential equations yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would it be alright if we kept working on my question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. From what we had about v' = 2y^(3) y', we see that (1/2)v' = y^(3) * y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now we can rewrite our original equation in terms of v's.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We get (1/2) t^2 * v' + 2t * v = 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are rewriting the equation in terms of v's wouldn't it be v=1 and then the orginal equation....you are intentfying the v as a variable given already.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i meant the orginal answer!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think you totally understand the question, ena, but thank you for trying to help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now we want to get rid of the coefficient on v' to be able to use the method of integrating factors easily.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Omnideus, do you know the method of integrating factors?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Kinda. I missed one of the lectures on it, so I am a little behind.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's basically what I tried to do when I first responded to your question. Here it's a little harder since the integrating factor isn't as obvious.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we have v'  (4/t) v = 2/t^2 after multiplying both sides by (2/t^2) to get rid of the coefficient on v'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is starting to look a little more approachable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the method of integrating factors, we want to express the left hand side as the derivative of the product of some function phi with the function v.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because the derivative of phi * v is phi * v' + phi' * v

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let me call the function phi, let me call it Y. Then we want to find some Y such that when we multiply Y to the Left Hand Side and get Yv'  (4/t) Y*v, we want that to equal Y v' + Y' v.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what we really want is some function Y such that (4/t)Y = Y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solving this, we get Y = t^(4), and this is the integrating factor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Multiplying both sides of v'  (4/t) v = 2/t^2 by t^(4), we see that the left hand side is the derivative of v*t^(4), while the right hand side is 2/t^6.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're pretty much done with solving for v, and once we solve for v, we've solved for y since v = y^(2).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When we have the left hand side is the derivative of v*t^(4) and the right hand side is 2/t^6, we can integrate both sides with respect to t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You think you got it from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I got an answer that looks correct. Did you do the check? It looks ugly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, but we can try wolfram alpha.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not really that tech savvy. Should I go to the site that other guy posted and you gave me the URL to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=t^2+*+y%27+%2B+2ty+%3D+y^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You might have written it a bit differently. http://www.wolframalpha.com/input/?i=v%27++%284%2Ft%29+v+%3D+2%2Ft^2 That's the solution to our equation for v. Then v = y^(2) gives our solution in terms of y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're a fluttering godsend, man. I would have struggled with that all night long and not made any progress.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can't thank you enough

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome. Thanks for letting me know I made a mistake. I'm gonna tell duongni now.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Good luck with your hw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, I hope duongni isn't in bed by now. THANKS A MILLION!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Other guy? hehee I am a girl! if you are refering to me for giving you the website.
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