[Orrin]

How did everyone else find this? If anyone else exists who took it. Ended up struggling on Q1 in finding the particular integral so left it and powered through Q2,3,4 instead which I thought were quiet straight forward.

[gannalise]

I did questions 1,3,4 What did you get for your y approximation in Euler's method??

[kingsingh]

-0.000677 i think

[Orrin]

What did you use for your PI for q1? I tryed (x^2)(e^-3x) and it didnt end up right, dont know if i made a mistake elsewhere or what. Seem to remember my first being small and positive and the final one small and negative, which seemed to make sense cos the number was passing pi, you get something like that?

[Orrin]

(Original post by kingsingh)
-0.000677 i think

Yeah that sounds very familiar!

[gannalise]

this is great news that's what i got!!! i got a different number each time so i was stressing out :L i used y=a(x^2)e(-3x) and it worked okay for me

[kingsingh]

(Original post by Orrin)
What did you use for your PI for q1? I tryed (x^2)(e^-3x) and it didnt end up right, dont know if i made a mistake elsewhere or what. Seem to remember my first being small and positive and the final one small and negative, which seemed to make sense cos the number was passing pi, you get something like that?

The PI was (x^2)(e^-3x), most of the terms should have cancelled out, you go on to work out that a = 0.5
Therefore y=(A+Bx)e^-3x +0.5x^2e^-3x

[Orrin]

(Original post by kingsingh)
The PI was (x^2)(e^-3x), most of the terms should have cancelled out, you go on to work out that a = 0.5
Therefore y=(A+Bx)e^-3x +0.5x^2e^-3x

Very good memory! I havent a clue what I did wrong, went over it a few times and just couldn't get it to work, kept losing a!

[gannalise]

i got a=0.5 too ! thank god. that was my answer

[gannalise]

also. how exactly did y=kx for infinitely many values of t because i just couldn't figure that one out > <

[kingsingh]

(Original post by gannalise)
also. how exactly did y=kx for infinitely many values of t because i just couldn't figure that one out > <

Do you remember your solutions for x and y?

[gannalise]

ummm. not masively. i think i got x=1+(Acost+Bsint)e^-2t and y=4-(Bcost-Asint)e^-2t.
My y one might not be exactly what i got but the x one is and i got A and B as 6 and 4

[Orrin]

(Original post by gannalise)
also. how exactly did y=kx for infinitely many values of t because i just couldn't figure that one out > <

Nore could I, ended up writing down anything i could think up in hope of getting a few marks haha

(Original post by gannalise)
ummm. not masively. i think i got x=1+(Acost+Bsint)e^-2t and y=4-(Bcost-Asint)e^-2t.
My y one might not be exactly what i got but the x one is and i got A and B as 6 and 4

I got that too!

[gannalise]

this is great news! i simplified it down to the form re(-3t)sin(t-alpha)=0 but i couldn't see how that worked for infinitely many values of t :L

[Orrin]

(Original post by gannalise)
this is great news! i simplified it down to the form re(-3t)sin(t-alpha)=0 but i couldn't see how that worked for infinitely many values of t :L

I remember the 4's cancelled out nicely, think i cancelled out the e's but then i was left with cosines and sines (cant remember any more detailed than that) that definitely didn't = 0 for every value of t haha. Wrote abit under as well proving it did for large values of t then realised that was just going back on myself, hopefully some working marks?

[gannalise]

yeah the es cancelled out and the fours. it was only 5 marks so i reckon we'd have probably got about 3 for finding k and whatnot.
i presume you got k as 4?

[RacingPro97]

For the y=kx question, you could use the particular solutions for x and y and rearrange to get an equation of the form ae^-3t(bsint+ccost)=0, and since e^-3t>0 for all t, then bsint+ccost=0. This gives tanx=-c/b, and since tant is periodic, there are infinitely many solutions for t such that y=kx (I think that k was equal to 4, so y=4x for infinitely mant t). I remember getting something like tant=-14/5, although this may not be correct because I don't remember much of it!

Did anyone get y=-x^2(1+cosx) for the first part of Q3? I think it is correct after checking on Wolfram Alpha, but I thought that the graph was a bit of a pain! The graph looks like oscillations of increasing amplitude, all below the x-axis, touching the x-axis at x=0,pi,3pi.

[gannalise]

oh right thank you i would never have got that :L

for question three the dy/dx+P(X)y=Q(X) style question? i don't remember anything like that. i think i got y=x^3+piX^2 so mine was just a simple curve O.o

[Orrin]

I did get k as 4, and very clever, quite a hard 2/3 marks!
For Q3 that sounds familiar, i know my graphs just how you described it, was a bit of a pain though because when I checked it in my calculator it was off the scale!

[gannalise]

i clearly did something terribly wrong on Q3 D: