Thread: Undetermined Coefficient question

1. Undetermined Coefficient question

(d-1)^2(d+1)y = 10cos2x

Work I've done:

yc:
1. (r-1)^2(r+1) = 0, r=1,1,-1
2. yc=(c1 +c2x)e^x + c3e^-x
3. Then expand the original: r^3 - r^2 - r + 1 = 10cos2x
4. So y^(3) - y" - y' +y = 10cos2x

How do I go about finding yp? yp=Acos2x + Bsin2x, and then find to the third derivative? Thanks,

JN

2. Originally Posted by Jim Newt
(d-1)^2(d+1)y = 10cos2x

Work I've done:

yc:
1. (r-1)^2(r+1) = 0, r=1,1,-1
2. yc=(c1 +c2x)e^x + c3e^-x
3. Then expand the original: r^3 - r^2 - r + 1 = 10cos2x
4. So y^(3) - y" - y' +y = 10cos2x

How do I go about finding yp? yp=Acos2x + Bsin2x, and then find to the third derivative? Thanks,

JN
$y_p=A\cos(2x)+B\sin(2x)$
$y_p'=-2A\sin(2x)+2B\cos(2x)$
$y_p''=-4A\cos(2x)-4B\sin(2x)$
$y_p^{(3)}=8A\sin(2x)-8B\cos(2x)$

Plug these into the equation to solve for A and B.

I hope this helps.

3. Originally Posted by TheEmptySet
$y_p=A\cos(2x)+B\sin(2x)$
$y_p'=-2A\sin(2x)+2B\cos(2x)$
$y_p''=-4A\cos(2x)-4B\sin(2x)$
$y_p^{(3)}=8A\sin(2x)-8B\cos(2x)$

Plug these into the equation to solve for A and B.

I hope this helps.
Thanks Empty Set, I originally came up with what you did. I reworked it and discovered another simple mistake in the arithmetic that screwed me up. Thanks,

JN