Old further maths special papers

Here's my solution which seems to disagree with the result expected in the question.

I cannot see anything wrong with your solution, so

either
there is a mistake (in those days I remember occasionally inside the papers where errata slips which of course has not survived to my hands)

or there is a silly mistake made and we both cannot see it

I will attempt the question from scratch later to see

thanks for pointing it out.

Here's my attempt at part (i). I'll put up part (ii) tomorrow:

I am sure many students are grateful to you, particularly because I have no marking schemes for pre 1987 and I am too busy to write solutions at present.

They won't be too grateful if I'm getting the answers wrong :-)

So where did I screw up in the mechanics question? I'm keen to see where my reasoning went wrong.

They won't be too grateful if I'm getting the answers wrong :-)

So where did I screw up in the mechanics question? I'm keen to see where my reasoning went wrong.

(please let me know of any obvious errors or typos)

Question.pdf

You're right. I was wrong. I failed to take into account that $u=v+a\omega$. This was rather obvious in retrospect. I get the same result as you when I make that change. I'll delete my original post as it may be misleading.

This is interesting. Can I post solutions to some of the questions if I have them?

I had a go at doing 1986 Q2, and after using the substitution given ($y = ve^x$), where v is a function of x,

I've arrived at

$e^x \left(2xv'' + v' \right) = 0 \implies 2xv'' + v' = 0$

I have not seen this form of 2nd Order ODE before, but is it possible to solve it with what I already know from FP2 Differential Equations?

I haven't looked at this question at all or checked your working, but note that if you define w = v' then you have a new 1st order DE 2xw' + w = 0 which you can solve using your standard integrating factor technique

I had a go at doing 1986 Q2, and after using the substitution given ($y = ve^x$), where v is a function of x,

I've arrived at

$e^x \left(2xv'' + v' \right) = 0 \implies 2xv'' + v' = 0$

I have not seen this form of 2nd Order ODE before, but is it possible to solve it with what I already know from FP2 Differential Equations?

I am afraid I cannot

the papers have been scanned into very large PDFs of around 5 Mb, and that is why I upload question by question.