You are Here: Home >< Maths

# Differential Equations watch

1. Ok im having trouble with some of these questions,
In question 2, im having trouble with part ii) i can find the auxillary equation and then find the general solution but it is not what the answer is, and i have no idea how to do part iv) in this question.
Also in question 3, i've managed to work my way through part i) and ii) but i have no idea about the rest of the questions, any tips or a worked method would be very helpful thank you
Attached Images

2. (2)
(ii)
General solution for x:
x = 2e^(-2t) + Ae^(-t) + Be^(2t)

It follows from (1) that

y
= (1/18)(11x + e^(-2t) - x')
= (1/18)[22e^(-2t) + 11Ae^(-t) + 11Be^(2t) + e^(-2t) - (-4e^(-2t) - Ae^(-t) + 2Be^(2t))]
= (1/18)[27e^(-2t) + 12Ae^(-t) + 9Be^(2t)]
= (3/2)e^(-2t) + (2/3)Ae^(-t) + (1/2)Be^(2t)

(iii)
Suppose that x -> 0 as t -> infinity. Then B = 0. So y -> 0 as t -> infinity.

Suppose that y -> 0 as t -> infinity. Then B = 0. So x -> 0 as t -> infinity.

(iv)
Assume that the solutions do not tend to 0. Then B 0.

As t -> infinity,

x ~ Be^(2t)
y ~ (1/2)Be^(2t)
y/x -> 1/2

--

Assume that the solutions do tend to 0. Then B = 0.

There are two cases to consider: (a) A = 0; (b) A 0.

In Case (a), x = 2e^(-2t) and y = (3/2)e^(-2t). So y/x -> 3/4 as t -> infinity.

In Case (b), as t -> infinity,

x ~ Ae^(-t)
y ~ (2/3)Ae^(-t)
y/x -> 2/3

So the possible limits are 3/4 and 2/3.
3. (Original post by geekypoo)
Also in question 3, i've managed to work my way through part i) and ii) but i have no idea about the rest of the questions, any tips or a worked method would be very helpful thank you
See attached.

Edit: messed up this q a bit. Got it fixed now (hopefully !!)
Attached Images

4. Thanks alot, Fermat what does the top line of the iii) part say? that might be the bit that im missing. Because i still dont understand it (sorry for not being able to read it)
5. Which module would this be (Edexcel)?

terminal velocity is Vt = 70 m/s
parachutisats velocity is v = 0.99*Vt
or
v/Vt = 0.99
v/70 = 0.99
=========
7. okies, thank you Fermat. it was just that one line the rest was fine.

And this is OCR (MEI) syllabus, Mechanics 4 (a.k.a Differential Equations)
8. right
9. (Original post by geekypoo)
okies, thank you Fermat. it was just that one line the rest was fine.

And this is OCR (MEI) syllabus, Mechanics 4 (a.k.a Differential Equations)
you can get the ocr past papers and MS/(examiner's reports) from here
passsword is: sirisaac
10. I know i can get the mark scheme and stuff, but i sometimes cant really see where some of the steps have some from, but when i see a fully worked version i can see where i have made mistakes or where i haven't done something simple.
I'll have another question posted if i cant get the worked answers from somewhere. But this time M3
By the way Fermat, in part iv) in the mark scheme it says that B= -59.5 not 69.4 like it says in your answer, is that just error carried forward throughout parts iv and v then?

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 13, 2005
The home of Results and Clearing

### 2,495

people online now

### 1,567,000

students helped last year
Today on TSR

### GCSE results day guidance

All you need to know is here

### University open days

1. Bournemouth University
Wed, 22 Aug '18
2. University of Buckingham
Thu, 23 Aug '18
3. University of Glasgow
Tue, 28 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams