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    (Original post by SeanFM)
    As in how \frac {x-1}{x+1} becomes 1 - \frac{2}{x+1}?
    Yes please, I know how you'd do it by long division but on the solution for the question they changed the numerator from (z-1) to (z+1) -2, so I was just wondering how you would use this technique please? Thanks
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    Can somone explain how:

    e^ int P dx dy/dx + ( e^ int P dx *Py) = e^ int P dx * Q

    d/dx( e^ int P dx *Q) = e^ int P dx *Q

    is the integral of e^ int P dx - Pe^ int P dx +C ?
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    Does anyone know whether partial fractions like those in question 18 here http://madasmaths.com/archive/maths_..._questions.pdf could be in the exam, not seem them with quadratic powers before! Thanks
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    Think I'll do the *** **** madas fp2 next Monday and Tuesday

    Sent from my SM-G925F using Tapatalk
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    (Original post by economicss)
    Yes please, I know how you'd do it by long division but on the solution for the question they changed the numerator from (z-1) to (z+1) -2, so I was just wondering how you would use this technique please? Thanks
    If we had \frac{x-1}{x+1} then we can 'add 0' to the numerator like so:

    \frac{x-1+(2-2)}{x+1}

    = \frac{(x+1)-2}{x+1} and then use that (a+b)/c = (a/c) + (b/c) to give \frac{x+1}{x+1} - \frac{2}{x+1}
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    (Original post by SeanFM)
    If we had \frac{x-1}{x+1} then we can 'add 0' to the numerator like so:

    \frac{x-1+(2-2)}{x+1}

    = \frac{(x+1)-2}{x+1} and then use that (a+b)/c = (a/c) + (b/c) to give \frac{x+1}{x+1} - \frac{2}{x+1}
    Thank you
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    Would anyone mind explaining the solution for this question? I cant understand how they have gone from the first line in the solution to the second. Thanks!
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    (Original post by lkara)
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    Would anyone mind explaining the solution for this question? I cant understand how they have gone from the first line in the solution to the second. Thanks!
    It's integrating factors. Basically the DE is an exact DE.
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    Does anyone know a site that has all of the IAL F2 papers? thanks very much
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Size:  504.5 KBCan anyone explain why for question one we don't differentiate the whole thing then put pi over 4 in ?in the mark scheme they just expanded cot then put then times it by the x - pi over 4. Just wondering why they didn't differiatiate both parts this time.
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    Attachment 542281542283

    (Original post by Tcoupe)
    Name:  image.jpg
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Size:  504.5 KBCan anyone explain why for question one we don't differentiate the whole thing then put pi over 4 in ?in the mark scheme they just expanded cot then put then times it by the x - pi over 4. Just wondering why they didn't differiatiate both parts this time.
    I just did the expansion of cotx (by differentiating and inputting pi/4), multiplied through by (x-(pi/4))

    Edit: Didn't mean to send it twice just useless with this phone
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    I'm trying to differentiate (dy/dx) = (e^-z) * (dy/dz). Using the product rule you know differentiating gives you (e^-z) * (the derivative of (dy/dz) with respect to x) + (dy/dz) * (the derivative of (e^-z) with respect to x) but I'm unsure how to differentiate something in terms of z with respect to x. I'd imagine it would be similar to implicit differentiation and I've gotten close to the right answer just playing around with the derivatives but I don't understand the maths behind it.

    Thanks for any help/explanation/tips and sorry for the formatting.
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    (Original post by Patrick2810)
    Does anyone know a site that has all of the IAL F2 papers? thanks very much
    There are only two but try this

    http://qualifications.pearson.com/en...nit%2FPaper-F2
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    (Original post by Xabier)
    I'm trying to differentiate (dy/dx) = (e^-z) * (dy/dz). Using the product rule you know differentiating gives you (e^-z) * (the derivative of (dy/dz) with respect to x) + (dy/dz) * (the derivative of (e^-z) with respect to x) but I'm unsure how to differentiate something in terms of z with respect to x. I'd imagine it would be similar to implicit differentiation.

    Thanks for any help/explanation/tips and sorry for the formatting.
    I think (someone correct me if I'm wrong) with the derivative of e^-z it's, like you said, similar to implicit differentiation so you get
    -e^-z (dz/dx).
    With the derivative of (dy/dz) with respect to x, you want to get an expression that means (d^2y/dz)*(d/dx) so I think it becomes (d^2y/dz^2)*(dz/dx)?
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    (Original post by PhysicsIP2016)
    I think (someone correct me if I'm wrong) with the derivative of e^-z it's, like you said, similar to implicit differentiation so you get
    -e^-z (dz/dx).
    With the derivative of (dy/dz) with respect to x, you want to get an expression that means (d^2y/dz)*(d/dx) so I think it becomes (d^2y/dz^2)*(dz/dx)?
    I think the derivative of (dy/dz) would just be (d^2y/dz^2)

    Or at least that's what I've been doing with the exam questions and it usually works out
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    (Original post by somevirtualguy)
    I think the derivative of (dy/dz) would just be (d^2y/dz^2)
    But it's with respect to x?
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    (Original post by PhysicsIP2016)
    But it's with respect to x?
    Oh my bad
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    (Original post by PhysicsIP2016)
    I think (someone correct me if I'm wrong) with the derivative of e^-z it's, like you said, similar to implicit differentiation so you get
    -e^-z (dz/dx).
    With the derivative of (dy/dz) with respect to x, you want to get an expression that means (d^2y/dz)*(d/dx) so I think it becomes (d^2y/dz^2)*(dz/dx)?
    Thanks, just checked and your right on both.
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    (Original post by Zacken)
    It's integrating factors. Basically the DE is an exact DE.
    Name:  working fp2 q.png
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    Thanks for your reply. I think I understand now. Would you mind checking the first few lines of my working to see if the logic follows?
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    (Original post by lkara)
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    Thanks for your reply. I think I understand now. Would you mind checking the first few lines of my working to see if the logic follows?
    Perfect.
 
 
 
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