Advanced Higher Maths 2012-2013 : Discussion and Help Thread

For questions 2 and 3 I expanded the whole thing out then found the coefficients. Is this the only method or is there a formula you can use to find the specific term?

I know this is used for two terms: $\sum\limits_{r=0}^n \displaystyle \binom{n}{r}x^{n-r}y^r$, but is there something for 3 in a bracket?

For questions 2 and 3 I expanded the whole thing out then found the coefficients. Is this the only method or is there a formula you can use to find the specific term?

Yes it's a different cycle.

Dude, quick question, to apply for a PhD, did you need to take the SAT?

Maths is NOT going well for anyone in our class.. (ps hi Joanne if you are creeping as usual) and I'm just reeeeally worried !

In our mock prelim, our scores were: 6%, 20%ish, 30%ish and 45%.
Really can't see anyone in our class passing! I know there is still plenty of time before the exam, but seriously struggling!

OH WELL, only 8 periods until we finish the course.

Uh, wow. In my class everyone passed (though there are only 5).

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Why are we not taught to do it this way.

Given that,
$x^2e^y \frac{dy}{dx}=1$

and y = 0 when x = 1, find y in terms of x.

$\int 0_y e^ydy= \int 1_x\frac{dx}{x^2}$
And so on.
Is it just me that finds this easier than the general solution method?

Well, I don't know much about the AH system but this one's just a straightforward separable differential equation...do you do integrating factors and the like?

Look at the edited version,
In the marking instructions for past papers,
we don't use limits, we always use the constant

Ah, I see. I thought you were just separating the variables first time :P

And you found that in the marking instructions? Integrating w.r.t. y and having a 'y' in your limit, and the same for 'x' doesn't make sense

Thats the same for my school. I think 3 people passed the prelim out of 18, and 2 of them were low fiftys Although we've got far more than 8 periods to finish the course We've got our Unit 2 nab on Thursday I'll probably try and power through whats left over easter tbh

Why are we not taught to do it this way.

Given that,
$\displaystyle x^2e^y \frac{dy}{dx}=1$

and y = 0 when x = 1, find y in terms of x.

$\displaystyle\int _0^y e^ydy= \int _1^x\frac{dx}{x^2}$
And so on.
Is it just me that finds this easier than the general solution method?

Aww I'm glad to hear we're not the only ones! thought we were all just **** haha! Your plan sounds a lot better than mine - ZERO motivation! We're starting the final outcome tomorrow ( )

Hello.

Struggling a bit with a question:



I get up to where you have to deduce the expressions, and then I'm stuck. The solutions say:



I don't get how you would know to add and subtract the two; am I forgetting a fundamental piece of simple theory?

Thanks.