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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use the method of integrating factors. Do you know how to do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y'+\frac{2}{x+4}y=(x+4)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh i haven't learned that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh  so, do you want me to just go through the motions and you look up the method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We look for a function mu(x) such that,\[\frac{d (\mu y)}{dx}=\mu (x+4)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0μ=2/(x+4) in this case?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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
 5 years ago
Best ResponseYou've already chosen the best response.0So here,\[\mu = \exp (\int\limits \frac{2}{x+4}dx)=e^{2\log (x+4)}=e^{\log (x+4)^2}=(x+4)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You then sub in mu, and integrate both sides:\[\frac{d((x+4)^2y)}{dx}=(x+4)^2(x+4)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\rightarrow (x+4)^2y=\int\limits_{}{}(x+4)^4dx\]\[=\frac{1}{5}(x+4)^5+c\]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
 5 years ago
Best ResponseYou've already chosen the best response.0I just checked with Wolfram and it's right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got everything besides those formulas haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You mean, why we set the problem up like that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Go to this page, http://www.khanacademy.org/ and look up 'integrating factors' under the differential equations section.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a math homework problem from my friend who is taking a different series of math courses here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh khanacademy. i watch their videos quite often on youtube :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm just sketching a proof...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for all this hot stuff :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you a math teacher ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I tutor tertiary and secondary maths to pay for postgrad.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In (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
 5 years ago
Best ResponseYou've already chosen the best response.0mu in mu(x) is just a symbol like f as in f(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you explain partial fractions? What i dont understand is that why f(x)/g(x) is equal to A1/(xr)+A2/(xr)^2+A3/(xr)^3+.......Am/(xr)^m if g(x) is a product of linear factors(xr)^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
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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 (xa)^m and thinking, "This is a polynomial of degree m, so why aren't I using a polynomial of degree m1 in my numerator?" Is that what it is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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)=(xa)^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
 5 years ago
Best ResponseYou've already chosen the best response.0for 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
 5 years ago
Best ResponseYou've already chosen the best response.0I see what you mean then should not it be like \[A \div (x+1) + Bx+c \div (x+1)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0Well, it depends on your original fraction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0You have a repeated linear factor and ONE irreducible quadratic. If the quadratic could be factored, THEN you'd have more LINEAR factors.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because (x+1)^2 is reducible . is that why the numerator there is B but not Bx+C?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's because you have a linear factor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0See, the whole point behind the procedure is to split up your original polynomial fraction into the sum of the most simplest fractions available.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why we actually do what we do is, like I said, covered in a modern algebra course.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0When 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh i watched most of their single variable calculus videos which are really helpful

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, just keep doing that. Then you won't need this site!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have you watched the video where the professor used blockstacking to explain convergence and divergence?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL my brain is kinda slow.. The more i read, the harder it is for me to think out of box

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh i watched it but i don't get it :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just keep mulling over it. You have to let your head put it together in its own time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://ocw.mit.edu/courses/mathematics/1801singlevariablecalculusfall2006/lecturenotes/lec36.pdf Do you explain how he gets the equation on page 7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you mind explaining*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0It doesn't. \[C_{n+1}=C_n+\frac{1}{n+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dichalao, I will be back in a min. Stay here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How you found it from that piles of questions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you see the picture i posted?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the professor said becasue the length of the block is 2, that is why C_(n+1)=Cn+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, this thing keeps freezing up.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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
 5 years ago
Best ResponseYou've already chosen the best response.0And each block has 2 units of length  don't forget that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the center of mass of each block individually is in the geometrical center in this situation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I added the geometry.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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
 5 years ago
Best ResponseYou've already chosen the best response.0The notes are what you should be paying attention to. They make complete sense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0C_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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0You build up a series,\[C_N=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{N}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the distance from the beginning to the center of mass of the Nth block.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0His 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 xdirection, 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
 5 years ago
Best ResponseYou've already chosen the best response.0But at the same time, the system has been set up to be stable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got what you saying here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you're fine with it?
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