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Edexcel FP2 - 2nd order differentiation question Watch

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    (a) Find the general solution of the differential equation
    4(d2y/dy2) - (dy/dx) - 3y = x^2

    [CF: Ae^((-3/4)x) + Be^x]


    (b) Find the particular solution for which, x=0, y=3 and dy/dx = 5

    according to the mark scheme, you start with y = px^2 + qx + r => y' = 2px + q, y'' = 2p

    anyone know why you have to assume that equation for y? does it have anything to do with the x^2 in the original equation (implying a quadratic in y)?
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    (Original post by the maths guy)
    (a) Find the general solution of the differential equation
    4(d2y/dy2) - (dy/dx) - 3y = x^2

    [CF: Ae^((-3/4)x) + Be^x]


    (b) Find the particular solution for which, x=0, y=3 and dy/dx = 5

    according to the mark scheme, you start with y = px^2 + qx + r => y' = 2px + q, y'' = 2p

    anyone know why you have to assume that equation for y? does it have anything to do with the x^2 in the original equation (implying a quadratic in y)?
    Yes, that's what I'd imagine the reasoning to be. The RHS is a quadratic and having y as a quadratic will also give you a quadratic on the LHS.
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    (Original post by Farhan.Hanif93)
    Yes, that's what I'd imagine the reasoning to be.
    When did you have your Cambridge interview btw?
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    (Original post by davidmarsh01)
    When did you have your Cambridge interview btw?
    The 8th.
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    (Original post by davidmarsh01)
    When did you have your Cambridge interview btw?
    Hey, how did your cam interview go?
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    (Original post by cpdavis)
    Hey, how did your cam interview go?
    Yeah, the interview went pretty well actually, but the test was worse Hope they liked me :p:
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    (Original post by Farhan.Hanif93)
    Yes, that's what I'd imagine the reasoning to be. The RHS is a quadratic and having y as a quadratic will also give you a quadratic on the LHS.
    so if the RHS was 3x, y would be ax + b?
    and if 2, y = ax?
 
 
 
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