Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

yeah

Answer 6

yeah

Answer 7

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

Answer 8

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

Answer 9

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.

Answer 10

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.

Answer 11

How should it look then?

Answer 12

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

Answer 13

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

Answer 14

any lucl?

Answer 15

*luck, even

Answer 16

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.

Answer 17

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.

Answer 18

that makes sense

Answer 19

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.

Answer 20

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.

Answer 21

would you be willing to walk me through that?

Answer 22

Sure.

Answer 23

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.

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

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

Answer 33

what's the url?

Answer 34

http://blog.wolframalpha.com/2009/12/01/step-by-step-math/

But I don't know if it can do differential equations yet.

Answer 35

fair enough

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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.

Answer 41

i meant the orginal answer!

Answer 42

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

Answer 43

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

Answer 44

Omnideus, do you know the method of integrating factors?

Answer 45

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

Answer 46

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.

Answer 47

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'.

Answer 48

this is starting to look a little more approachable

Answer 49

Good. Haha.

Answer 50

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.

Answer 51

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

Answer 52

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.

Answer 53

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

Answer 54

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

Answer 55

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.

Answer 56

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

Answer 57

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.

Answer 58

You think you got it from there?

Answer 59

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

Answer 60

No, but we can try wolfram alpha.

Answer 61

kay

Answer 62

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

Answer 63

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

Answer 64

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.

Answer 65

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

Answer 66

I can't thank you enough

Answer 67

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

Answer 68

Good luck with your hw

Answer 69

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

Answer 70

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