How would I use the Bernoulli Equation to solve t^2dy/dt+2ty-y^3=0, t>0?

- anonymous

How would I use the Bernoulli Equation to solve t^2dy/dt+2ty-y^3=0, t>0?

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- anonymous

The 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

Thank you very, very much, verifry. You're answer to duongi's curiously identical question was immensely helpful.

- anonymous

I'm having a really hard time with the check on this one. would you be willing to walk me through it?

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## More answers

- anonymous

You mean checking the solution to see if it solves the original equation?

- anonymous

yeah

- anonymous

yeah

- anonymous

I keep ending up with a -3ty^2dy/dt that doesn't fit

- anonymous

would you be willing to walk me through that part as well?

- anonymous

You 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

The 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

How should it look then?

- anonymous

Let me work it out and check it this time before typing anything.

- anonymous

cool beans. Thank you again for sticking with me through all of this. I really appreciate the help!

- anonymous

any lucl?

- anonymous

*luck, even

- anonymous

I'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

The 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

that makes sense

- anonymous

- anonymous

So 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

would you be willing to walk me through that?

- anonymous

Sure.

- anonymous

Starting 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

So we get t^2 * (y'/y^3) + 2t / y^2 = 1.

- anonymous

Now 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

Normally, after solving lots of these problems and generalizing the method, you'd just let v = y^(1-n) 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

Why can't we use v=y^(1-n)?

- anonymous

We 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^(1-n) might seem to pop out of nowhere.

- anonymous

I sort of read through the website explanation, but I just don't pick stuff up very well that way. We could use v=y^(1-n) though

- anonymous

Yeah let's let v = y^(1-n). 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

If you aren't sure of the answer you should visit aplha math help its on google, and shows you the exact answer!

- anonymous

Wow, alpha math help is pretty cool. I didn't know about that either.

- anonymous

what's the url?

- anonymous

http://blog.wolframalpha.com/2009/12/01/step-by-step-math/
But I don't know if it can do differential equations yet.

- anonymous

fair enough

- anonymous

would it be alright if we kept working on my question?

- anonymous

yeah. From what we had about v' = -2y^(-3) y', we see that (-1/2)v' = y^(-3) * y'

- anonymous

Now we can rewrite our original equation in terms of v's.

- anonymous

We get (-1/2) t^2 * v' + 2t * v = 1.

- anonymous

if 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

i meant the orginal answer!

- anonymous

I don't think you totally understand the question, ena, but thank you for trying to help.

- anonymous

Now we want to get rid of the coefficient on v' to be able to use the method of integrating factors easily.

- anonymous

Omnideus, do you know the method of integrating factors?

- anonymous

Kinda. I missed one of the lectures on it, so I am a little behind.

- anonymous

It'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

So 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

this is starting to look a little more approachable

- anonymous

Good. Haha.

- anonymous

In 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

Because the derivative of phi * v is phi * v' + phi' * v

- anonymous

Let 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

So what we really want is some function Y such that
-(4/t)Y = Y'

- anonymous

Solving this, we get Y = t^(-4), and this is the integrating factor.

- anonymous

Multiplying 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

We're pretty much done with solving for v, and once we solve for v, we've solved for y since v = y^(-2).

- anonymous

When 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

You think you got it from there?

- anonymous

Yeah, I got an answer that looks correct. Did you do the check? It looks ugly.

- anonymous

No, but we can try wolfram alpha.

- anonymous

kay

- anonymous

I'm not really that tech savvy. Should I go to the site that other guy posted and you gave me the URL to?

- anonymous

http://www.wolframalpha.com/input/?i=t^2+*+y%27+%2B+2ty+%3D+y^3

- anonymous

You 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

You're a fluttering godsend, man. I would have struggled with that all night long and not made any progress.

- anonymous

I can't thank you enough

- anonymous

You're welcome. Thanks for letting me know I made a mistake. I'm gonna tell duongni now.

- anonymous

Good luck with your hw

- anonymous

haha, I hope duongni isn't in bed by now. THANKS A MILLION!

- anonymous

Other guy? hehee I am a girl! if you are refering to me for giving you the website.

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