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    (the highlighted part) I don't know how to find (+x+2) the rest I got correct..where did I mess up? I know the " x " and " 2b " was missed out but it was for a reason

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    (Original post by ckfeister)
    (the highlighted part) I don't know how to find (+x+2) the rest I got correct..where did I mess up? I know the " x " and " 2b " was missed out but it was for a reason
    I haven't looked through this in great detail, and this may be what you mean when you talk about the 2b thing, but shouldn't your trial function be of the form a cos(2x) + b sin(2x) + cx + d? That extra constant might make all the difference when you differentiate twice and substitute back in.
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    (Original post by ckfeister)
    (the highlighted part) I don't know how to find (+x+2) the rest I got correct..where did I mess up? I know the " x " and " 2b " was missed out but it was for a reason

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Views: 9
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    If you let y=a\sin 2x +b\cos 2x + cx+d and then solve for all 3 constants I believe you'll get your answer. I might be wrong though. Try it out and see what you get.
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    (Original post by ckfeister)
    (the highlighted part) I don't know how to find (+x+2) the rest I got correct..where did I mess up? I know the " x " and " 2b " was missed out but it was for a reason
    I've seen it through now, and it is the lack of a constant term in your trial function that is the problem. Try it again with one there.
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    (Original post by Pangol)
    I've seen it through now, and it is the lack of a constant term in your trial function that is the problem. Try it again with one there.
    but wouldn't d=0 as there is no constant on the other side?
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    if you have x on the right then you need to include ax + b in your Particular Integral
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    (Original post by ckfeister)
    but wouldn't d=0 as there is no constant on the other side?
    If you include d from the beginning, you end up with a constant term on the LHS of d - 2c. It is this that is equal to zero, and since c is not zero, neither is d.

    In general, of you have a quadratic of degree n on the RHS, then your trial function also has to include a quadratic of degree n. There is one exception to this - if the differential equation only has terms in the second and first derivative, and no term in y, you instead need to use a quadratic of degree n+1 (although this time, curiously, it does not have to have a constant term!).
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    (Original post by Pangol)
    I've seen it through now, and it is the lack of a constant term in your trial function that is the problem. Try it again with one there.
    your right, thx

    (Original post by Desmos)
    If you let y=a\sin 2x +b\cos 2x + cx+d and then solve for all 3 constants I believe you'll get your answer. I might be wrong though. Try it out and see what you get.
    it was cx+d
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    Two things to point out:

    - As others have said, in your trial function, you need the term: cx + d,
    so your trial function should be:

     y(x) = a \sin(2x) + b \cos(2x) + cx + d, (where a, b, c and d are constants).

    - You are using b twice in your trial function, as you are using (as your trial function):

     y(x) = a \sin(2x) + b \cos(2x) + b.
 
 
 
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