I would help, although probably later on in the evening as I'm a bit busy atm.(Original post by Mr Dangermouse)
If I post a list of questions could anyone help me? Wish I could offer more than rep
Advanced Higher Maths 20112012 : Discussion and Help
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(Original post by Mr Dangermouse)
If I post a list of questions could anyone help me? Wish I could offer more than rep 
(Original post by Mr Dangermouse)
If I post a list of questions could anyone help me? Wish I could offer more than rep 
(Original post by Mr Dangermouse)
If I post a list of questions could anyone help me? Wish I could offer more than rep 
I guess I'm in for some maths on a Friday night.

dy/dx + y/x = x
Anybody? 
Then: is your integrating factor. Multiply both sides of the equation by this integrating factor and see if you can think where to go from there.
The LaTex isn't clear but that's e raised to the integral of P(x).
If you get stuck.Spoiler:ShowP(x) = 1/x. So e raised to the integral of 1/x which is (as you know) lnx. e^lnx = x. Then multiply both sides by x. 
Erm... wut?

I normally solve them by splitting the variables.

(Original post by Mr Dangermouse)
I normally solve them by splitting the variables.
Steps for the Integrating factor method:
1. Ensure the equation is in standard form  dy/dx + p(x)y = Q(x)
2. Identify P(x) and
3. Write out the integrating factor
4.Multiply both sides by the integrating factor
5. The LHS will be
6. Integrate both sides.
7. Substitute in initial conditions/rearrange to make y the subject. 
(Original post by Hype en Ecosse)
So it's a first order linear differential equation of form:
What's P(x)? What's Q(x)?
You can then use P(x) to find the integrating factor that's of the form , which you multiply the entire original equation by, surely you can remember where to go from here?
My solution:
Spoiler:Show
(Original post by Mr Dangermouse)
I normally solve them by splitting the variables. 
(Original post by Mr Dangermouse)
I normally solve them by splitting the variables.
dy/dx = xy
And you'd make it dy/y = xdx and integrate. But you don't have that here, have you done firstorder linear differential equations?
(Original post by JordanR)

I don't know if they have a specific name actually. I don't think I've been told one if they do.
But if they're like: then we can split the variables up.
(Original post by Hype en Ecosse)
c is a completely arbitrary constant, so I thought one didn't need to divide the c? 
(Original post by JordanR)
I don't know if they have a specific name actually. I don't think I've been told one if they do.
But if they're like: then we can split the variables up.
Hence why it's semantics! Although it would become important if one was given an initial condition and one was also asked to find the value of one's constant, so it's definitely good practice for one to do so. 
(Original post by Hype en Ecosse)
c is a completely arbitrary constant, so I thought one didn't need to divide the c? 
(Original post by TheUnbeliever)
Yes, but they're not the same arbitrary constant, so you need to either show the division or rename it. Same reason . It's not just semantics: your last step implies c = 0.
This was shortly after I got a row about "remember that she's done a degree and you've not." So... 
I hate it when teachers try and argue from authority without justifying it at all.

Hmm, this is quite a nice question. Quite a few different techniques required for it:
Solve by use of standard integrals.
Also, I have a few questions:
What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?
Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure). 
(Original post by Quintro)
Hmm, this is quite a nice question. Quite a few different techniques required for it:
Solve by use of standard integrals.
Also, I have a few questions:
What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?
Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).
Yes, I'd think so. So if you had (x+y)^5 you'd find the term that only had y in it.
Probably, but I've never seen them in any past papers. Not looked at many though.