This looks like a very simple one, but i ran out of ideas. I do not get the answer and unlikely will get it without some help:

y'' + 16y = 24 cos(4x)

I found homogeneous solution:

y = c

I haven't re-written it using euler's formula yet.

So, for particular solution the general form is:

y = x(Acos(4x) + Bsin(4x))

I differentiated it and plugged into the DE and here's what i have:

-8Asin(4x) + 8Bcos(4x) + xsin(4x)*[16A - 16B] + xcos(4x)*[16B - 16A] = 24cos(4x);

It seems like last two terms need to go away, but then A = B and to get rid of the sin A = 0! the answer is totally different though.

Where am i going wrong?

Thanks for any hints.

EDIT: or would y

I'm confused... i can't work it out with x

y'' + 16y = 24 cos(4x)

I found homogeneous solution:

y = c

_{1}e^{i4}+ c_{2}e^{-i4}I haven't re-written it using euler's formula yet.

So, for particular solution the general form is:

y = x(Acos(4x) + Bsin(4x))

I differentiated it and plugged into the DE and here's what i have:

-8Asin(4x) + 8Bcos(4x) + xsin(4x)*[16A - 16B] + xcos(4x)*[16B - 16A] = 24cos(4x);

It seems like last two terms need to go away, but then A = B and to get rid of the sin A = 0! the answer is totally different though.

Where am i going wrong?

Thanks for any hints.

EDIT: or would y

_{p}= x^{2}(Acos(4x) + Bsin(4x))I'm confused... i can't work it out with x

^{2}either ...
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