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Grex > Reality > #30: a tricky little d.e. for you mathmaticians | |
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| Author |
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diznave
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a tricky little d.e. for you mathmaticians
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Nov 2 10:04 UTC 1996 |
any math buffs out there...i'm having a bit of a time with this problem
its a nasty little d.e. problem:
4y"+36y=csc(3x)
the trick is that i have to solve it w/ out the annihilator method and w/ out
u.c. sets...i've been trying to use variation of parameters but to no avail
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| 11 responses total. |
orinoco
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response 1 of 11:
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Nov 26 00:49 UTC 1996 |
umm...err...sound of brains popping
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rlawson
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response 2 of 11:
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Dec 22 14:35 UTC 1996 |
You know, there is a math conference. :)
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orinoco
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response 3 of 11:
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Dec 22 19:49 UTC 1996 |
well, the login here does mention mathematics, so...
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rlawson
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response 4 of 11:
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Dec 22 22:55 UTC 1996 |
That much I know, but it entails mathematics as a mean by which to understand
the universe.
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orinoco
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response 5 of 11:
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Dec 23 01:36 UTC 1996 |
details, details...
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ahtina
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response 6 of 11:
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Jan 2 07:07 UTC 1997 |
by the way... what is 'csc'?
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eskarina
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response 7 of 11:
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Jan 2 07:25 UTC 1997 |
cosecant, or 1/(sinx). Its one of those truly useless things, I've never
understood why they can't just write (sin x) ^ (-1) in all the theorems they
found out they can write with the trig stuff by doing things to the equation
(sin x) ^ 2 + (cos x) ^ 2 = 1, such as dividing thru by (sin x) ^ 2, so you
get 1 + (cot x) ^ 2 = (csc x) ^ 2.
Ugly stuff. I hated trig, and precalc.
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ahtina
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response 8 of 11:
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Jan 3 10:58 UTC 1997 |
Oh cool... we generally use 'cosec'... now lemme try this problem....
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eprom
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response 9 of 11:
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Dec 17 02:02 UTC 2007 |
ok...I know this is an old problem but i'll take a stab at it...
For the general characteristic equation we need to find the roots
4R^2+36R=0 and solving for R gives you +/- 3i thus Yc(x) = C1 cos(3x) + C2
sin(3x)
now if you let Yc take the form Yc = C1*y1(x) + C2*y2(x) you can solve for
Yp(x) by using the formula:
Yp(x) = -y1(x) int( y2(x)*f(x) / W)dx + Y2(x) int( y1(x)*f(x) / W)dx
where W is the wronskian of y1(x) and y2(x):
W = det([cos(3x), sin(3x) ; diff(cos(3x),x) , diff(sin(3x),x)]) => 3
now using the Yp(x) formula from above when I evaluate the integral on my ti-89
I get:
Yp(x) = [-x*cos(3x)] / 3 + [sin(3x)*ln(sin(3x))] / 9
so the complete solution is y(x) = Yc(x) + Yp(x) or:
C1 cos(3x) + C2 sin(3x) + [-x*cos(3x)] / 3 + [sin(3x)*ln(sin(3x))] / 9
I could easily see how anyone could get lost in the trig functions if they
didn't have a ti-89 and had to use the tables.
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eprom
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response 10 of 11:
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Dec 17 02:08 UTC 2007 |
hmm....using Maple I get:
C1*cos(3x)+C2*sin(3x)sin(3*x)+(1/36)*ln(sin(3*x))*sin(3*x)-(1/12)*x*cos(3*x)
but using Matlab I get:
sin(3*t)*C2+cos(3*t)*C1+1/(-36+144*cos(x)^2)/sin(x)
I'm thinking the Matlab anwser is wrong since there should be s 3*x term inside
the last two trig functions.
For some reason the Maple answer multiplies my Yp solution through by (1/4).
I'm doubting myself now....I dunno if my answer in the previous response was
correct.
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cmcgee
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response 11 of 11:
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Dec 17 19:41 UTC 2007 |
*watches carefully*
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