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C4 (Not MEI) - Thursday June 21 2012, PM

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    (Original post by wibletg)
    Yeah

    Does anyone else think they give a tonne of marks for solving differential equations?
    Why the hell have they made the C3/C4 book differential equations misc questions so hard? Annoying to say the least

    Well anyway as promised, several hours of going through all the misc questions (for C4). I NEED HELP ON THESE
    (They're all questions from the misc section btw)
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    Can i trouble anyone here for the jan 12 paper? i can't find it anywhere!
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    (Original post by Doctor.)
    Why the hell have they made the C3/C4 book differential equations misc questions so hard? Annoying to say the least

    Well anyway as promised, several hours of going through all the misc questions (for C4). I NEED HELP ON THESE
    (They're all questions from the misc section btw)
    1B and 1E, I think it wants you to use Chain rule.
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    (Original post by Choppyy)
    1B and 1E, I think it wants you to use Chain rule.
    That's what I thought, but I don't understand why we can't simply rearrange 1B and then differentiate it bit by bit? :eek:
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    (Original post by Doctor.)
    That's what I thought, but I don't understand why we can't simply rearrange 1B and then differentiate it bit by bit? :eek:
    There's no reason you can't.
    If you try both ways you could use it to verify sin(2x)=2sinxcosx

    For 11. b) let x be \pi - \theta now remember that sin\theta= sin(\pi - \theta) and -cos \theta = cos(\pi- \theta).
    Also expand sin(2x) before making teh substitution and then simplify again.
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    (Original post by Killjoy-)
    There's no reason you can't.
    If you try both ways you could use it to verify sin(2x)=2sinxcosx

    For 11. b) let x be \pi - \theta now remember that sin\theta= sin(\pi - \theta) and -cos \theta = cos(\pi- \theta).
    Also expand sin(2x) before making teh substitution and then simplify again.
    Thank you so much, I feel like a huge idiot after yoy explained 11B :lol: I got it now

    Understand both 1b & 1e too so thanks

    Just them other few to go and I think C4 has gone well
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    (Original post by Doctor.)
    Thank you so much, I feel like a huge idiot after yoy explained 11B :lol: I got it now

    Understand both 1b & 1e too so thanks

    Just them other few to go and I think C4 has gone well
    8b is a funny one, unless I'm being an idiot.

    It's going to be difficult to split that 0.5x from the right side of the differential equation...
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    (Original post by wibletg)
    8b is a funny one, unless I'm being an idiot.

    It's going to be difficult to split that 0.5x from the right side of the differential equation...
    At first I thought maybe I could just move it like its a normal addition/subtraction. That was obviously wrong, so I spent a wile trying to unpick and get the terms over to their side to integrate.

    In the end, I gave up :/
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    For 9 \displaystyle\int \dfrac{y}{1-y^2}dy=\displaystyle\int dt + c

    If you adjust the LHS using a constant of -2 it will be of the form \displaystyle\int\dfrac{f^{'}(y)  }{f(y)}dy which integrates to ln(f(y))
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    (Original post by Killjoy-)
    For 9 \displaystyle\int \dfrac{y}{1-y^2}dy=\displaystyle\int dt + c

    If you adjust the LHS using a constant of -2 it will be of the form \displaystyle\int\dfrac{f^{'}(y)  }{f(y)}dy which integrates to ln(f(y))
    So it Integrated would be -2ln(1-y^2)?

    thank you
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    (Original post by wibletg)
    8b is a funny one, unless I'm being an idiot.

    It's going to be difficult to split that 0.5x from the right side of the differential equation...
    (Original post by Doctor.)
    At first I thought maybe I could just move it like its a normal addition/subtraction. That was obviously wrong, so I spent a wile trying to unpick and get the terms over to their side to integrate.

    In the end, I gave up :/
    The question said verify. So I presume this is OK:
    Differentiate the second expression they have given wrt to t.
    Then rearrange the second expression (undifferentiated form) so that 1.4t is the subject and sub into the first equation.
    After a little simplification you will find that they are the same.
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    (Original post by Doctor.)
    So it Integrated would be -2ln(1-y^2)?

    thank you
    Close, but no cigar -1/2ln(1-y^2)

    Sorry I just realised I said a constant of -2 earlier
    I meant -1/2
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    (Original post by Doctor.)
    So it Integrated would be -2ln(1-y^2)?

    thank you
    With these questions (multiple things on the denominator) you'll either want it in the form \displaystyle\int\dfrac{f^{'}(x)  }{f(x)}dx or you'll be using a substitution. If the power of x is one less on the top, then it's the former. It might involve doubling the fraction (as in the above case) to get it in that form, meaning you have to half it as well to make sure you've not mathematically changed it. Substitutions tend to be trigonometric.

    I've got a tablet now, so might be doing some diagrams in response to questions
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    Received my C4 mock result today (Jan 2012) and got 71/72! Obviously delighted but would prefer that result on June 21st!
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    (Original post by Choppyy)
    Received my C4 mock result today (Jan 2012) and got 71/72! Obviously delighted but would prefer that result on June 21st!
    The Jan '12 paper wasn't actually that bad! I hope it's similar to ours though
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    http://www.ocr.org.uk/download/pp_08..._jan_l_gce.pdf
    Im in need of some help with q9 part 2... I have done the first part fine , but I cant even show working for part 2 because I dont know where to start ..
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    Anybody??
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    (Original post by wibletg)
    Hey all, good to see some familiar faces sitting this

    Not finding it too difficult at the minute, having done M3 differential equations are second nature.

    Only thing I'm not keen on is quotient and remainder as I hate the long division method
    I avoid the long division method at all costs. I like some weird reverse grid method that reminds me of GCSE :') it's basically equating the coefficients, but in a grid
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    Right you lot, tomorrow is my day to focus on C4 so I will be running through all the misc questions (again). I will be posting up some tricky problems probably around lunch so please do help! In the end, its helping you really!!!

    Where else should I do questions from? Other than past papers and the big book?
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    (Original post by Smiley Face :))
    Anybody??
    They used integration by parts straight away and let u=lnx as when differentiated it is 1/x * and that works out quite nicely.
    *I think that's a standard method which may come up often.

    I tried substitution, it isn't the method they use but it works.
    I let x=e^{-y} (You will see why I used y soon) and ended up with \displaystyle\int^{ln\frac{1}{3}  }_0 ye^y\ dy.

    Then I applied integration by parts, like before which function is v and which is \frac{du}{dx} is very important. (This also depends on which side of the parts equation you rearrange to but anyway...)
    If I used u instead of y for the letter of my substitution earlier then it would get very confusing at this stage with u's and v's :P

    Their alternative method involved substitution, but they ended up with the definite integral \displaystyle\int^{ln3}_0 te^{-t}\ dt instead.
    (And then they followed by using integration by parts.)

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Updated: June 22, 2012
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