Solve the differential equation y'+2y/(x+4)=(x+4)^2
where y=2 when x=0. y(x)= ?

- anonymous

Solve the differential equation y'+2y/(x+4)=(x+4)^2
where y=2 when x=0. y(x)= ?

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- anonymous

You can use the method of integrating factors. Do you know how to do that?

- anonymous

\[y'+\frac{2}{x+4}y=(x+4)^2\]

- anonymous

Oh i haven't learned that

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- anonymous

Oh - so, do you want me to just go through the motions and you look up the method?

- anonymous

yea sure

- anonymous

We look for a function mu(x) such that,\[\frac{d (\mu y)}{dx}=\mu (x+4)^2\]

- anonymous

μ=2/(x+4) in this case?

- anonymous

This is one of those things I usually derive for myself each time (since there's less chance of stuffing up), but for an equation in the form\[y'+p(x)y=q(x)\]the integrating factor is\[\mu(x)=e^{\int\limits_{}{}p(x)dx}\]

- anonymous

So here,\[\mu = \exp (\int\limits \frac{2}{x+4}dx)=e^{2\log (x+4)}=e^{\log (x+4)^2}=(x+4)^2\]

- anonymous

You then sub in mu, and integrate both sides:\[\frac{d((x+4)^2y)}{dx}=(x+4)^2(x+4)^2\]

- anonymous

\[\rightarrow (x+4)^2y=\int\limits_{}{}(x+4)^4dx\]\[=\frac{1}{5}(x+4)^5+c\]That is,\[(x+4)^2y=\frac{1}{5}(x+4)^5 +c \rightarrow y = \frac{1}{5} (x+4)^3+\frac{c}{(x+4)^2}\]

- anonymous

I just checked with Wolfram and it's right.

- anonymous

I got everything besides those formulas haha

- anonymous

You mean, why we set the problem up like that?

- anonymous

yes

- anonymous

Yeah, that's why I asked if you were familiar with the method, because I knew it wouldn't make sense (looks like you're pulling something out of nowhere).

- anonymous

Go to this page,
http://www.khanacademy.org/
and look up 'integrating factors' under the differential equations section.

- anonymous

this is a math homework problem from my friend who is taking a different series of math courses here

- anonymous

oh khanacademy. i watch their videos quite often on youtube :)

- anonymous

I'm just sketching a proof...

- anonymous

I'll scan.

- anonymous

thanks for all this hot stuff :)

- anonymous

are you a math teacher ?

- anonymous

I tutor tertiary and secondary maths to pay for postgrad.

- anonymous

Oh

- anonymous

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- anonymous

In (1) we just multiply through by some function, mu(x). We know nothing about it yet. Then we put it aside and consider d(mu y)/dx. Then we link the two together to force a mu that has the property we want.

- anonymous

mu in mu(x) is just a symbol like f as in f(x)

- anonymous

yes

- anonymous

\[\mu = \mu (x)\]

- anonymous

it makes sense :)

- anonymous

Good :)

- anonymous

Can you explain partial fractions? What i dont understand is that why f(x)/g(x) is equal to A1/(x-r)+A2/(x-r)^2+A3/(x-r)^3+.......Am/(x-r)^m if g(x) is a product of linear factors(x-r)^m
Why doesn't the degree of the numerators increase as the degree of the denominator increases, but rather stays as constants A1....Am?

- anonymous

The theory behind it is that you have repeated roots in the polynomial g(x). The method has to be modified to take account of the repeating of the roots.

- anonymous

I think you're asking this because you see the situation where there is a quadratic in the denominator sometimes, and you're told to write the numerator as Ax+B, say. You're looking at (x-a)^m and thinking, "This is a polynomial of degree m, so why aren't I using a polynomial of degree m-1 in my numerator?"
Is that what it is?

- anonymous

Yea exactly

- anonymous

If so, it's because the quadratics (and higher polynomials) in those cases are irreducible over the field we're working in.
For example, in the real field, \[x^2+x+1\]has no real factors. So over the real field, this thing is *irreducible*: we cannot write it as a product of linear factors. We have to then use a numerator that has a polynomial of degree one less.
In the case where g(x)=(x-a)^m, if you saw it in expanded form, although you wouldn't necessarily be able to eyeball it, it *is* reducible into linear factors.
That's all, really.

- anonymous

for instance, if \[f(x)\div g(x) = (3x)\div (x+1)^3 -> A \div (x+1)+B \div (x+1)^2+ C \div (x+1)^3 \] the degree of denominator is increasing, but the degree of the numerator is not. which seems to contradict with Ax+b/ (x^2+x+c)

- anonymous

OH

- anonymous

I see what you mean then should not it be like \[A \div (x+1) + Bx+c \div (x+1)^2\]

- anonymous

The actual theory requires a study of modern algebra. You start talking about polynomials over fields and all sorts of things.
The procedure is easy and necessary to do a lot of practical problems, which is why 'method' is taught over 'reason'.

- anonymous

Well, it depends on your original fraction.

- anonymous

Oh I see it.

- anonymous

No, you have *linear* factors here, so you'd write\[\frac{3x}{(x+1)^3}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}\]

- anonymous

If you had something like\[\frac{3x}{(x+1)^2(x^2+x+1)}\]then you'd write\[\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+x+1}\]

- anonymous

You have a repeated linear factor and ONE irreducible quadratic.
If the quadratic could be factored, THEN you'd have more LINEAR factors.

- anonymous

because (x+1)^2 is reducible . is that why the numerator there is B but not Bx+C?

- anonymous

It's because you have a linear factor.

- anonymous

See, the whole point behind the procedure is to split up your original polynomial fraction into the sum of the most simplest fractions available.

- anonymous

Why we actually do what we do is, like I said, covered in a modern algebra course.

- anonymous

I like to discover what is behind the methods rather than just plugging numbers into formulas haha it is ok if it takes too much effort to explain :)

- anonymous

When you come across your problem, you consider these things when deciding on the decomposition:
1) are any of my factors repeated?
2) are any of my polynomials irreducible (you're really only ever going to see quadratics in a course you do)?
When setting up the decomp., you ALWAYS use a polynomial in the numerator that is 1 degree LESS than the polynomial in the denominator.
If your polynomial is all raised to some power, you have to go through the repeating process; e.g.\[\frac{3x}{(x^2+x+1)^2}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{(x^2+x+1)^2}\]

- anonymous

Yeah, like I said, you have to do a course in modern algebra.
There's nothing stopping you looking at lecture notes or something. MIT Open Courses sometimes have good material.

- anonymous

Oh i watched most of their single variable calculus videos which are really helpful

- anonymous

Yeah, just keep doing that. Then you won't need this site!

- anonymous

Have you watched the video where the professor used block-stacking to explain convergence and divergence?

- anonymous

No...

- anonymous

LOL my brain is kinda slow.. The more i read, the harder it is for me to think out of box

- anonymous

Oh i watched it but i don't get it :(

- anonymous

Just keep mulling over it. You have to let your head put it together in its own time.

- anonymous

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/lecture-notes/lec36.pdf
Do you explain how he gets the equation on page 7?

- anonymous

do you mind explaining*

- anonymous

I could explain it if I had more time...I have to get ready for uni/work. I will definitely look at it and get back to you.

- anonymous

Thanks :)

- anonymous

Later ;)

- anonymous

I had a look at your question during the day. I'll write something up for you soon. The lecturer just made some assumptions in the end that they didn't make explicit, mainly because (I'm assuming) that part of the lecture wasn't core.

- anonymous

I watched the lecture video too, i don't get why \[C_{n+1}=C_{n}+1 \] he says the center of mass of the N+1 block(the last block) is 1 unit further than the center of mass of the N blocks

- anonymous

It doesn't. \[C_{n+1}=C_n+\frac{1}{n+1}\]

- anonymous

Dichalao, I will be back in a min. Stay here.

- anonymous

How you found it from that piles of questions

- anonymous

E-mail

- anonymous

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- anonymous

Can you see the picture i posted?

- anonymous

yes

- anonymous

the professor said becasue the length of the block is 2, that is why C_(n+1)=Cn+1

- anonymous

Sorry, this thing keeps freezing up.

- anonymous

No problem :)

- anonymous

Yeah, the center of the (n+1)st block is at C_n+1 because you're measuring horizontally from the tip of the first block on top to the center of the (n+1)st block. I'll draw it.

- anonymous

And each block has 2 units of length - don't forget that.

- anonymous

and the center of mass of each block individually is in the geometrical center in this situation.

- anonymous

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- anonymous

I added the geometry.

- anonymous

You're just dealing in distance. The C_n is the collective distance up to the beginning of the (n+1)st block and the 1 comes from the 2 units of block.
He's not talking about the center of mass, just geometric center I think.

- anonymous

The notes are what you should be paying attention to. They make complete sense.

- anonymous

C_1 is one unit away from the origin, C_2=1+1/2, C_3=1+1/2+1/3,, then when it comes to C_(n+1)=Cn+1 because every succeeding term is getting bigger and bigger, but the component added to the previous term is getting smaller and smaller

- anonymous

Yes, basically. If you look at the notes, you'll see that C_n is the collective distance (hence why I think he's used 'C') from the beginning of the first block up to the center of mass of the nth block.
The physics is what determines these distances. The mathematics is what determines the solution.
Here, if you look, you have
C_0 = 0 (because it's at the tip, and therefore no distance)
C_1 = C_0 + 1 (because the center of mass of a block is in its geometric center)
C_2 = C_0 + C_1 +1/2 (because the physics says this is where the CoM is and we're aiming to place the center of mass at the edge of the block below)
C_3=C_0+C_1+C_2+1/3 (because of what I just said for C_2)
and so it continues...

- anonymous

You build up a series,\[C_N=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{N}\]

- anonymous

Following so far

- anonymous

for the distance from the beginning to the center of mass of the Nth block.

- anonymous

His point in this exercise is to show that this physical system can be set up in theory AND that you can make its horizontal projection (i.e. how the blocks extend in the x-direction, say) as large as you want (people would tend to think it would be impossible).
He shows you can make it as large as you want BECAUSE the (collective) distance is
\[C_N=\sum_{n=0}^{N}\frac{1}{n}\]...a harmonic series.
These things DON'T converge, so as n gets larger, so does the distance stretched...that's his point.

- anonymous

But at the same time, the system has been set up to be stable.

- anonymous

I got what you saying here

- anonymous

So you're fine with it?

- anonymous

Sure :) Thanks~~

- anonymous

No probs. :)

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