MEI Differential Equations - June 12th 2014

I think it was x=Ae^t + Be^2t +1/12e^-2t

Thought it was an alright paper. What did everyone put for the last part of Q2 when you had to comment on the revised model.

Only thing I wasn't sure on Q1 was when one of the roots of the auxiliary equation was 0, did that mean that you just added a constant onto the end of the C.F, since I figured that Ke^0 = K?

Think I got the same as 'Mandragoran'

For question 1, with the third order equation when you had to graph it, did you get it tend to 2?

I got it tended to 12...

I got it tended to 12...

I was afraid of that, my friend did aswell if anyone remembers the question could you explain why?

and i just redid that question and it gave you the graph y=10-10cosx which means it tends to 10 not 12 surely? unless i wrote out the conditions wrong

Right I was an idiot, on question one (i), some how I thought root 4 over 2 was 2 (exam nerves) and it meant I got -1+-2j and as the original right hand side had 30cos2t I multiplied Asin2t and Bcos2t in my particular solution both by x and tried to solve, this question was only worth 9 marks, do you think I would get any method marks? so sad and angry

I think you will get all (or at least most of) the method marks, for 2 reasons:
1. It's unfair of them to deduct all the method marks for one arithmetic error at the beginning of a question, as it is no fault of method, rather just a human error. (i.e They would follow through from your -1+-2j for most of the question.)
2. You showed that you knew what you were doing with multiplying by x, that shows good knowledge of method, meaning they will give you the benefit of the doubt on the method marks.