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FP2: Quick question about 2nd order differential equations watch

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    I was studying 2nd order differential equations in the textbook and I don't understand when they say "the complementary function includes a constant term A the particular integral cannot also be the constant" etc (the example was d^2y/dx^2 - 2dy/dx = 3).
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    (Original post by creativebuzz)
    I was studying 2nd order differential equations in the textbook and I don't understand when they say "the complementary function includes a constant term A the particular integral cannot also be the constant" etc (the example was d^2y/dx^2 - 2dy/dx = 3).
    The complementary function is y=A+Be^2x
    If you choose a PI as y=λ, notice that this solves the D.E (y''-2y'=0) as it is part of the CF. So, λ cannot be a PI. Instead, you take the next 'most' complicated function, that is y=λx as the PI
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    (Original post by Gome44)
    The complementary function is y=A+Be^2x
    If you choose a PI as y=λ, notice that this solves the D.E (y''-2y'=0) as it is part of the CF. So, λ cannot be a PI. Instead, you take the next 'most' complicated function, that is y=λx as the PI
    I thought y=A+Be^2x is for equations which have two equal roots? But aren't these roots "two real distinct roots" so shouldn't the complementary function be y=Ae^ax + Be^bx
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    (Original post by creativebuzz)
    I thought y=A+Be^2x is for equations which have two equal roots? But aren't these roots "two real distinct roots" so shouldn't the complementary function be y=Ae^ax + Be^bx
    y=(A+Bx)e^2x is for repeated roots. Yes you do have the correct CF for the two real distinct roots but one of them is zero hence y=A+Be^2x.
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    (Original post by Matt#)
    y=(A+Bx)e^2x is for repeated roots. Yes you do have the correct CF for the two real distinct roots but one of them is zero hence y=A+Be^2x.
    Oh I see! But when finding "the roots" when do you use y=e^mx and when do you factorise? Because in this case, if you factorise, you could've ended up with (m-3)(m+1), hence I was confused about how you managed to get y=A + Be^2x
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    (Original post by creativebuzz)
    Oh I see! But when finding "the roots" when do you use y=e^mx and when do you factorise? Because in this case, if you factorise, you could've ended up with (m-3)(m+1), hence I was confused about how you managed to get y=A + Be^2x
    For a differential equation of the form ay'' + by' + cy = f(x)

    The auxiliary equation is an^2 + bn + c = 0

    That's the quadratic you solve to get the complimentary function. The constant term does not form the quadratic.
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    (Original post by Matt#)
    For a differential equation of the form ay'' + by' + cy = f(x)

    The auxiliary equation is an^2 + bn + c = 0

    That's the quadratic you solve to get the complimentary function. The constant term does not form the quadratic.
    Err I'm still a little confused about when to use y=e^mx and when to factorise?
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    (Original post by creativebuzz)
    Oh I see! But when finding "the roots" when do you use y=e^mx and when do you factorise? Because in this case, if you factorise, you could've ended up with (m-3)(m+1), hence I was confused about how you managed to get y=A + Be^2x
    The Auxiliary equation you have should be m^2-2m = 0

    which factorises to m(m-2) = 0.

    You've formed your AE as m^2 - 2m - 3 = 0 which is wrong.
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    (Original post by creativebuzz)
    Oh I see! But when finding "the roots" when do you use y=e^mx and when do you factorise? Because in this case, if you factorise, you could've ended up with (m-3)(m+1), hence I was confused about how you managed to get y=A + Be^2x
    The equation you have is:

    y'' - 2y' = 3. The homogeneous version of the equation (i.e. only having terms involving y, y' and y'') is

    y'' - 2y = 0.

    It is this equation that you need to form an aux equation from and then factorize.

    [It looks as if you are treating y''-2y=3 as if it were the equation y'' - 2y = 3y, at least in terms of what you think the auxiliary equation is].
 
 
 
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