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# Second order differentiaton watch

1. (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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2. (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.
3. (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

Attachment 695884695886
Attachment 695884695886696000
[attach]6.958846958866959e+23[/attach]
[attach]6.958846958866958e+29[/attach]
If you let 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.
4. (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.
5. (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?
6. if you have x on the right then you need to include ax + b in your Particular Integral
7. (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!).
8. (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.

(Original post by Desmos)
If you let 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
9. 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:

(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):

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