Sin+cos solution undetermined coeff. de

[EvLer]

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 = c₁eⁱ⁴ + c₂e^-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 = x²(Acos(4x) + Bsin(4x))
I'm confused... i can't work it out with x² either ...

 

[rickphysics]

I find the last two terms do cancel

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

y' = Acos(4x) + Bsin(4x) + x (-4Asin(4x) + 4Bcos(4x))

y" = -4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

Putting back into the equation y" + 16y = 24cos(4x) gives

-4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

+ 16 ( x(Acos(4x) + Bsin(4x)) ) = 24 cos(4x)

Here we are left with

-8Asin(4x) + 8Bcos(4x) = 24 cos(4x)

so B = 3 and A = 0

so a

y(p) = 3x sin(4x)

 

[EvLer]

Thanks for reply.
that was my "other" solution. Good thing it matched yours. Come to find out: the book answer was wrong