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    Polar Coordinates is probably the most difficult subject in FP2, anyone have any guides or strategies to learn it?
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    Gonna go ahead and bump this because people seem to have forgotten about it
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    Revising this at the moment, doing differential equations, first order subs.
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    Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations
    Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question

    Here's the question if anyone wants to have a go.

    Show that the substitution \displaystyle x=e^{z} transforms the differential equation


    x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)
    into the equation


    \dfrac{d^{2}y}{dz^{2}}+\dfrac{dy  }{dz}-2y = 3z\ \ \ \ \ \ - (II)

    Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
    (14 marks)
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    (Original post by edothero)
    Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations
    Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question

    Here's the question if anyone wants to have a go.

    Show that the substitution \displaystyle x=e^{z} transforms the differential equation


    x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)
    into the equation


    \dfrac{d^{2}y}{dz^{2}}+\dfrac{dy  }{dz}-2y = 3z\ \ \ \ \ \ - (II)

    Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
    (14 marks)
    This is a nice question, started a minute ago and I'm struggle
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    (Original post by kkboyk)
    This is a nice question, started a minute ago and I'm struggle
    Hint: \dfrac{dy}{dx} =\ ? \times ?

    Using z
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    (Original post by edothero)
    Hint: \dfrac{dy}{dx} =\ ? \times ?

    Using z
    I had to revisit 2nd ODE to know which form of particular integral my answer must have. But it took the piss.
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    (Original post by edothero)
    Had an FP2 mock, messed up on one of the complex numbers transformation questions (question was worth 6 marks - should get at least 2). I hate transformations
    Apart from that I'm fairly confident I got the rest correct. Including the last question, which was a beautiful second order question

    Here's the question if anyone wants to have a go.

    Show that the substitution \displaystyle x=e^{z} transforms the differential equation


    x^{2}\dfrac{d^{2}y}{dx^{2}} + 2x\dfrac{dy}{dx} - 2y = 3lnx\ \ \ \ \ \ - (I)
    into the equation


    \dfrac{d^{2}y}{dz^{2}}+\dfrac{dy  }{dz}-2y = 3z\ \ \ \ \ \ - (II)

    Hence find the general solution to the differential equation (II) and obtain the general solution of the differential equation (I). Giving your answer in the form y=f(x).
    (14 marks)
    Might as well. We have: \frac{\mathrm{d}x}{\mathrm{d}z} = e^z so since e^z \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}z} , we then have, by taking second derivatives:

    and \frac{d^2y}{\mathrm{d}x^2} = (e^{-z} y'' - y'e^{-z})\frac{\mathrm{d}z}{\mathrm{d}  x} = e^{-2z}(y'' - y' )

    So our DE becomes:

    \displaystyle y'' - y' + 2y' - 2y = 3z \iff y'' + y' - 2y = 3z as required.

    Now this is easily solvable, the auxiliary quadratic is \alpha^2 + \alpha - 2 = 0 \Rightarrow \alpha = -2, 1

    So complementary solution is Ae^{-2z} + Be^{z} and the particular solution -\frac{3}{2}, so:

    \displaystyle y = Ae^{-2z} + Be^{z} - \frac{3z}{2} - \frac{3}{4}, then

    \displaystyle y = \frac{A}{x^2} + Bx - \frac{3\ln x}{2} - \frac{3}{4}
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    (Original post by Zacken)
    ...
    Yep very nice
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    (Original post by kkboyk)
    I had to revisit 2nd ODE to know which form of particular integral my answer must have. But it took the piss.
    What have you got up to? You might want to look at Zacken's solution. If you don't understand anything I can elaborate
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    (Original post by Zacken)
    So complementary solution is Ae^{-2z} + Be^{z} and the particular solution -\frac{3}{2}, so:

    \displaystyle y = Ae^{-2z} + Be^{z} - \frac{3}{2}, then

    \displaystyle y = \frac{A}{x^2} + Bx - \frac{3}{2}
    Hang on, just realised, your particular solution is incorrect
    Spoiler:
    Show
    y = \mu z + \alpha

    \dfrac{dy}{dz} = \mu

    \dfrac{d^{2}y}{d^{2}z} = 0

    \therefore \mu - 2\mu z - 2\alpha = 3z

    Split this up

    -2\mu = 3 \ \ \ \ \ \rightarrow \ \ \ \ \ \mu = -\dfrac{3}{2}

    \mu - 2\alpha = 0 \ \ \ \ \  \rightarrow \ \ \ \ \ \alpha = -\dfrac{3}{4}

    \therefore y = \dfrac{A}{x^{2}} + Bx - \dfrac{3}{2}lnx - \dfrac{3}{4}
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    (Original post by edothero)
    Hang on, just realised, your particular solution is incorrect
    Spoiler:
    Show
    y = \mu z + \alpha

    \dfrac{dy}{dz} = \mu

    \dfrac{d^{2}y}{d^{2}z} = 0

    \therefore \mu - 2\mu z - 2\alpha = 3z

    Split this up

    -2\mu = 3 \ \ \ \ \ \rightarrow \ \ \ \ \ \mu = -\dfrac{3}{2}

    \mu - 2\alpha = 0 \ \ \ \ \  \rightarrow \ \ \ \ \ \alpha = -\dfrac{3}{4}

    \therefore y = \dfrac{A}{x^{2}} + Bx - \dfrac{3}{2}lnx - \dfrac{3}{4}
    Lol yeah, that's what I get for doing it in my head thanks for pointing it out.
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    (Original post by edothero)
    Hang on, just realised, your particular solution is incorrect
    Spoiler:
    Show
    y = \mu z + \alpha

    \dfrac{dy}{dz} = \mu

    \dfrac{d^{2}y}{d^{2}z} = 0

    \therefore \mu - 2\mu z - 2\alpha = 3z

    Split this up

    -2\mu = 3 \ \ \ \ \ \rightarrow \ \ \ \ \ \mu = -\dfrac{3}{2}

    \mu - 2\alpha = 0 \ \ \ \ \  \rightarrow \ \ \ \ \ \alpha = -\dfrac{3}{4}

    \therefore y = \dfrac{A}{x^{2}} + Bx - \dfrac{3}{2}lnx - \dfrac{3}{4}
    Yeah I realised my mistake here thanks
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    Hi, please could someone explain how to do question 8c on this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf in particular how you get arc sin 3/13 as I'm not sure how this is found! Thanks
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    (Original post by economicss)
    Hi, please could someone explain how to do question 8c on this paper https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf in particular how you get arc sin 3/13 as I'm not sure how this is found! Thanks
    this should help, find the angle from the origin to the point (5,12) then use the info on the diagram to find the min and max tangents

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    (Original post by DylanJ42)
    this should help, find the angle from the origin to the point (5,12) then use the info on the diagram to find the min and max tangents

    Thank you so much!
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    Show that the locus of \displaystyle z=x+yi, \arg\left(\frac{z-3i}{z-5}\right)=\frac{\pi}{4}, has the cartesian equation \displaystyle (x-1)^2+(y+1)^2=17, subject to \displaystyle 5y+3x<15.
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    I think this is a little bit above the syllabus, so don't worry if you don't know how to do it. Also I may have made an error.
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    (Original post by EricPiphany)
    Show that the locus of \displaystyle z=x+yi, \arg\left(\frac{z-3i}{z-5}\right)=\frac{\pi}{4}, has the cartesian equation \displaystyle (x-1)^2+(y+1)^2=17, subject to \displaystyle 5y+3x<15.
    Spoiler:
    Show
    I think this is a little bit above the syllabus, so don't worry if you don't know how to do it. Also I may have made an error.
    Nice questions
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    (Original post by Joshthemathmo)
    Nice questions
    Thanks
    The question is actually in the Edexcel textbook, 3F 7 d, as a 'sketch' problem, I just decided to try it algebriacally.
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    (Original post by EricPiphany)
    Thanks
    The question is actually in the Edexcel textbook, 3F 7 d, as a 'sketch' problem, I just decided to try it algebriacally.
    Fair enough, i haven't really thoroughly gone through the textbook questions. It's nice to see problems tackled from different positions
 
 
 
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