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

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

Oh i haven't learned that

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

yea sure

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

μ=2/(x+4) in this case?

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

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

I just checked with Wolfram and it's right.

I got everything besides those formulas haha

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

yes

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

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

I'm just sketching a proof...

I'll scan.

thanks for all this hot stuff :)

are you a math teacher ?

I tutor tertiary and secondary maths to pay for postgrad.

Oh

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

yes

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

it makes sense :)

Good :)

Yea exactly

OH

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

Well, it depends on your original fraction.

Oh I see it.

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

It's because you have a linear factor.

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

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

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

No...

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

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

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

do you mind explaining*

Thanks :)

Later ;)

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

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

How you found it from that piles of questions

E-mail

Can you see the picture i posted?

yes

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

Sorry, this thing keeps freezing up.

No problem :)

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

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

I added the geometry.

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

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

Following so far

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

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

I got what you saying here

So you're fine with it?

Sure :) Thanks~~

No probs. :)