- #1
EvLer
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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 yp = x2(Acos(4x) + Bsin(4x))
I'm confused... i can't work it out with x2 either ...
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 yp = x2(Acos(4x) + Bsin(4x))
I'm confused... i can't work it out with x2 either ...
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