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# OCR MEI DE June 17 2016

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1. Hello, just did the MEI de exam and wanted to see how people found it. How did you find it?
2. Erm, I remeber for one of the soultions in Q2: y= 1/[4-3sin(x)], the last part of Q1: y= (2+t)e^-0.5t + 4e^-3t I think is 2 what I got - If you sort of remind me what the questions were, I think it'll jog my memory. I think for Question 4 the coefficients were 1/2 and zero.
3. THat was the easiest exam of my life. ssaros sorry to inform you but im certain you got that question wrong, my father is a maths teacher and he confimed that the coefficients were infact 1/4 and -1/4
4. (Original post by SolomonAsghar)
THat was the easiest exam of my life. ssaros sorry to inform you but im certain you got that question wrong, my father is a maths teacher and he confimed that the coefficients were infact 1/4 and -1/4
Are you sure Solomon, I can't work out how you would have gotten that.
5. ANSWERS I GOT FOR Q1,2,3.
Question 1: second part particular solution was:x=4*e^-0.5t+3/2*e^-1.5t+0.5*e^0.5tStationary point at ln(9) (not sure if this is for the first or second DE that we solved....) Particular solution for the last part was:x=(4+t)*e^-0.5t+2*e^-1.5t.
Question 2: First part particular solution was y=x^3(xsinx + cosx -pi/2)next part solution was y= -1/(3sinx -4) greatest value of y is 1 as bottom of fraction has a "greatest" value (value closest to 0 that would make top biggest) of -1. -1/-1 =1. Euler's method gave 0.010775... =0.0108. as the y value when x =1 (step length of 0.01.
Question 3: First part was a show that. Use N2L. integreate answer of part i to get the answer to part ii (was again a show that so easy to know if its correct). Distance travelled when v=48 was 352.5m. Sketch the graph, tends to v=sqrt(2450) from below. For the next part use N2l to get a differential equation with v and t. Solving this with the initial conditions t=0 when v=48 gives: v=1.25*(g+143/5*e^-0.8t). To find the distance travelled between t=0 and t=5, integrate the answer we just obtained between these limits gives: 105.12m.Didnt even bother looking at question 4 cos i hate it. Hope it went well
6. (Original post by Rawsonj)
Question 1: second part particular solution was:x=4*e^-0.5t+3/2*e^-1.5t+0.5*e^0.5tStationary point at ln(9) (not sure if this is for the first or second DE that we solved....) Particular solution for the last part was:x=(4+t)*e^-0.5t+2*e^-1.5t.
Question 2: First part particular solution was y=x^3(xsinx + cosx -pi/2)next part solution was y= -1/(3sinx -4) greatest value of y is 1 as bottom of fraction has a "greatest" value (value closest to 0 that would make top biggest) of -1. -1/-1 =1. Euler's method gave 0.010775... =0.0108. as the y value when x =1 (step length of 0.01.
Question 3: First part was a show that. Use N2L. integreate answer of part i to get the answer to part ii (was again a show that so easy to know if its correct). Distance travelled when v=48 was 352.5m. Sketch the graph, tends to v=sqrt(2450) from below. For the next part use N2l to get a differential equation with v and t. Solving this with the initial conditions t=0 when v=48 gives: v=1.25*(g+143/5*e^-0.8t). To find the distance travelled between t=0 and t=5, integrate the answer we just obtained between these limits gives: 105.12m.Didnt even bother looking at question 4 cos i hate it. Hope it went well
I'm pretty sure I got the same answers as you, can't remember what I got the first part of question one though. But yeah, everything else seems fine.
7. Did anyone do question 4??
Did anyone do question 4??
Yeah what did u get ?
9. I got all cos and sins in my answer, for the last part i got 2pi/3
I got all cos and sins in my answer, for the last part i got 2pi/3
Yup i got that as well. Although i wrote it as 2.09 3sf cos its a time.
Yup i got that as well. Although i wrote it as 2.09 3sf cos its a time.
Can you remember anything else you got?? Trying to remember but ive forgotten it all already hahaha😂😂
12. (Original post by Dumbledoge)
I'm pretty sure I got the same answers as you, can't remember what I got the first part of question one though. But yeah, everything else seems fine.
Did it say y=1 was a maximum or minimum??
Did it say y=1 was a maximum or minimum??
Maximum
14. I got 4pi/3 for last part of 4 :/ Did you solve the quadratic in sine to get
sin(t) = 1/2 , -1 ?

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Did anyone do question 4??
Yeah, I didn't get 2pi/3 but rather got pi/3... should get method marks though if I'm wrong. When you formed the 2nd order DE for x did you not have a dx/dt term either? I just got complex roots
16. (Original post by magicdarts)
Yeah, I didn't get 2pi/3 but rather got pi/3... should get method marks though if I'm wrong. When you formed the 2nd order DE for x did you not have a dx/dt term either? I just got complex roots
yeah my equation was d2x/dt2 + 4x = 0 so my roots were +- 2j
so my answer only had sins and cos in it, no exponentials
17. (Original post by Geo_)
I got 4pi/3 for last part of 4 :/ Did you solve the quadratic in sine to get
sin(t) = 1/2 , -1 ?

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yes i did, i initially got 4pi/3 too but when i checked on my graphical calculator the answer was exactly half this. what you had to do was find the values of t, so the primary values gave you -pi/2 and pi/6.
i assume you did what i did first and got the secondary value which was 3pi/2 then did this take pi/6 to give 4pi/3
if you work out the other secondary value you get 5pi/6, if you do this take pi/6, or do pi/6 - -pi/2, you get 2pi/3
you will still get method marks though, i only imagine one mark will be lost, dont worry!!
18. Because of last part of q4, will it be 89 for 100 or still 90/90 for 100 ums ?
Because of last part of q4, will it be 89 for 100 or still 90/90 for 100 ums ?
june 13 was 89
jan 13 was 87
jan 12 was 88

icant remember the standard of those papers but was ours harder or easier? its possible for 100 ums not to be 90 but who knows haha
yes i did, i initially got 4pi/3 too but when i checked on my graphical calculator the answer was exactly half this. what you had to do was find the values of t, so the primary values gave you -pi/2 and pi/6.
i assume you did what i did first and got the secondary value which was 3pi/2 then did this take pi/6 to give 4pi/3
if you work out the other secondary value you get 5pi/6, if you do this take pi/6, or do pi/6 - -pi/2, you get 2pi/3
you will still get method marks though, i only imagine one mark will be lost, dont worry!!
Yeah that's precisely what I did. Ok that's alright then, I can definitely cope with 1 mark lost.
Although, because it asked for the time difference between the 1st and 2nd times they are equal, surely that will be the difference between when t=pi/6 and t=3pi/2 ?

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