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

11 responses total.

umm...err...sound of brains popping

You know, there is a math conference. :)

well, the login here does mention mathematics, so...

That much I know, but it entails mathematics as a mean by which to understand
the universe.

details, details...

by the way... what is 'csc'?

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.

Oh cool... we generally use 'cosec'... now lemme try this problem.... 

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.

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.

*watches carefully*

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