[Meteorshower]

Ok, I'm starting to revise properly for this now, and I can't be the only! This thread can be used to resolve maths problems before the exam and discuss the exam afterwards.

Now I have a problem

Question 6 in the 2002 paper

Use the substitution x+2 = 2tan(theta) (i'll call theta k to save myself the bother of typing it) to solve integral of (1/(x^2+4x+8))dx

I completed the square to make the denominator (x+2)^2 +4

x + 2 = 2tank
dx = 2/(1+k^2)

Substituting in I get integral of ((1+k^2)/(8tan^2k+8))dk

I've simplified the denominator to 8sec^2k but i'm not sure how that helps. I'm probably missing something obvious here or made some stupid mistake but help will be appreciated

[ej49]

I think you've differentiated you subsitution wrong. from there, i'm getting myself mixed up with this question too!

If you have x + 2 = 2tank
tank = (x+2)/2
K = tan^-1 ((x+2) +2)
then differentiate and get dk/dx
and i think you then get 2/(x+2)^2 +1

the difficulty with this is everyone is taught different methods! so its hard to see where people are coming from... hmm

[Meteorshower]

I think you're right, I shall try and finish this off now! Thanks



Edit - Your method was right, but dk/dx = 2/((x+2)^2 + 4)

I got it anyway, thanks for the help.

[Bobby92]

That question was in our prelim. It was a bit of a nightmare I had to waste 20 minutes on it until it clicked.

What do we all think will come up?

Transformation matrices haven't appeared in a while, and there wasn't an inductive proof in last years paper, so I'd say that was fair game.

[TheUnbeliever]

IMHO:

Particular solution of a 2nd order non-homogeneous ODE.
Euclidean algorithm to solve a simple Diophantine equation.

[Meteorshower]

I hope there is an absolutely beastly 2nd order non homogeneous ODE, I like them I'd rather a difficult question on that than something else. Not too fussed though.

[spoony2oo8]

the thing i dont like the most in AH is asymptotes.
i really should start studying :-/

[rbnphlp]

I got a differnt answer.
1/(x+2)^2+4
if i integrate that wouldnt i get :
ln((x+2)^2+4)/2(x+2)

and then sub 2tanx=x+2( theta=x)
I get,
ln(4tan^2 x+4)/4tanx
taking 4 out
=ln(4)sec^2(x)/4tanx
=2ln(2secx)/4tanx
=ln(2secx)/2tanx

[TheUnbeliever]

The brilliant thing (from the point of view of getting marks, rather than pedagogically!) about the 2nd order non-homogeneous ODEs is that there's a method, so you have 10 easy marks requiring very little insight.

[lynseyx]

Well, I'm freaking out about AH Maths after failing the 2nd prelim. However that may be a good thing as I find I always remember the stuff I got wrong in prelims.

I absolutely cannot do things like "show that z=3+3i is a root of the equation z^3-18z+108=0 and obtain the remaining roots" and discovered in the prelim that I'm useless at interchanging trig things, you know, when you see something containing sec/tan/sin/cosec/cot or whatever and have to somehow realise it's actually a complicated version of cos^2. Sorry, no idea if you'll understand what I'm finding difficult. This is why I'm not studying maths at uni, lol

[Meteorshower]

Originally Posted by rbnphlp

I got a differnt answer.
1/(x+2)^2+4
if i integrate that wouldnt i get :
ln((x+2)^2+4)/2(x+2)

and then sub 2tanx=x+2( theta=x)
I get,
ln(4tan^2 x+4)/4tanx
taking 4 out
=ln(4)sec^2(x)/4tanx
=2ln(2secx)/4tanx
=ln(2secx)/2tanx

You need to sub back in x for theta though.

I'll call theta k again to help confusion

2tank = x + 2

tank = (x + 2)/2

k = tan^-1((x + 2)/2)

If you've done your working right this should give you an end result of (1/2)((tan^-1((x+2)/2)) + c

I can't be bothered to check if your way is correct also right now though XD

[Meteorshower]

Originally Posted by TheUnbeliever

The brilliant thing (from the point of view of getting marks, rather than pedagogically!) about the 2nd order non-homogeneous ODEs is that there's a method, so you have 10 easy marks requiring very little insight.

And if we're going to be cruel about it, as far as i'm aware they aren't done very well, so that's a lot of people missing out on 10% keeping the grade boundary comfortably low

[TheUnbeliever]

Originally Posted by Meteorshower

And if we're going to be cruel about it, as far as i'm aware they aren't done very well, so that's a lot of people missing out on 10% keeping the grade boundary comfortably low

Don't you have anything better to do than read the examiner's reports?

[Meteorshower]

Originally Posted by TheUnbeliever

Don't you have anything better to do than read the examiner's reports?

In my defence I haven't for absolutely ages I just remember them well!

[Ape Gone Insane]

Originally Posted by TheUnbeliever

The brilliant thing (from the point of view of getting marks, rather than pedagogically!) about the 2nd order non-homogeneous ODEs is that there's a method, so you have 10 easy marks requiring very little insight.

W T F ?

I have learned a few new words today.

[TheUnbeliever]

Originally Posted by Ape Gone Insane

W T F ?

I have learned a few new words today.

ODE = Ordinary Differential Equation. Trivial differential equations are in Higher, as I recall. The 'ordinary' distinguishes it from partial differential equations (PDEs): ODEs are composed with standard, total derivatives (the type you cover in school), PDEs are composed with partial derivatives. The order of a differential equation is the greatest power of the differential operator in the equation (for example, an equation including but no greater derivative would have order 2).

A partial derivative tells you the rate of change of a function of multiple variables with respect to a single one of those variables, with all others held constant (for example, the volume of a cone is a function of the radius of the circular base, and of its height; the partial derivative will tell you how quickly the volume of the cone changes if you keep one of those constant and vary the other). You won't cover PDEs until university.

Homogeneous appears to mean that the solutions are all linear transforms of each other, but you can take it to mean equations of the form (as opposed to non-homogeneous where the right hand side is a non-zero function of x).

[Lachlan]

Wouldn't mind a non-homogeneous ODE personally. Did that 2002 paper today, what did you get on it?

Got 65% myself, pretty pleased just need to bump up my mark another 10-15% to be safe.

[Meteorshower]

Originally Posted by Lachlan

Wouldn't mind a non-homogeneous ODE personally. Did that 2002 paper today, what did you get on it?

Got 65% myself, pretty pleased just need to bump up my mark another 10-15% to be safe.

I never do them properly, I just go through it and look something up if I don't understand it. It was only really the question I posted and that series question that bothered me too much though I think.

[Lachlan]

Also I got a different solution actually:

dx = 2sec^2k dk

therefore:

=2sec^2k dk/4tan^2k + 4
=2sec^2k dk/4(tan^2k + 1)
=2sec^2k dk/4sec^2k
=1/2 dk
=1/2k + C
=tan^-1((x+2)/2)/2 + C

[Meteorshower]

That's pretty good, didn't think of that