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Bilal
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#1
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#1
I'm calling it hard because my teacher didn't know how to do it. :rolleyes:

neway, could someone help me on it plz:

Show that the function y = cos(kx) / x, where x =/ (does not equal) 0 and k is a non-zero constant, satisfies the differential equation (d^2)y/dx^2 + 2/x. dy/dx + (k^2)y = 0.
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noether
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#2
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just differentiate it using the product or quotient rule, and then plug the results into the equation. it works for me, its just too messy and long for me to put it up
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evariste
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#3
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(Original post by Bilal786)
I'm calling it hard because my teacher didn't know how to do it. :rolleyes:

neway, could someone help me on it plz:

Show that the function y = cos(kx) / x, where x =/ (does not equal) 0 and k is a non-zero constant, satisfies the differential equation (d^2)y/dx^2 + 2/x. dy/dx + (k^2)y = 0.
xy=cos(kx)
y+xdy/dx=-ksin(kx)
dy/dx+dy/dx+xd2y/dx2=-k^2coskx
xd2y/dx2+2dy/dx+k^2coskx=0
d2y/dx2+2/xdy/dx+k^2coskx/x=0
d2y/dx2+2/xdy/dx+k^2y=0
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Fermat
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look below,

(Original post by Nima)
y = cos(kx) / x = cos(kx).x^(-1)
dy/dx = -cos(kx)/x^2 - ksin(kx)/x = -cos(kx)/(x^2) - kxsin(kx)/(x^2) = [-cos(kx) - kxsin(kx)]/(x^2) = -[cos(kx) + kxsin(kx)]/(x^2)

dy/dx = -[cos(kx) + kxsin(kx)]/(x^2) = -[cos(kx) + kxsin(kx)].(x^(-2))
d2y/dx^2 = - d/dx [cos(kx) + kxsin(kx)].(x^(-2)) = -{[cos(kx) + kxsin(kx)].(-2x^-3) + (x^-2).[k^2xcos(kx) - ksin(kx)] = -{-[cos(
...
= 0?

mmmm…
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BCHL85
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(Original post by Nima)
y = cos(kx)/x = x^-1.cos(kx)
dy/dx = (-x^-2)cos(kx) + (x^-1)(-ksin(kx)) = -[cos(kx)/x^2] - ksin(kx)/x
= -[cos(kx)/x^2] - kxsin(kx)/x^2 = [-cos(kx) - kxsin(kx)]/x^2 = -[cos(kx) + kxsin(kx)]/x^2 = -[cos(kx) + kxsin(kx)][x^-2]

d2y/dx^2 = - d/dx [cos(kx) + kxsin(kx)][x^-2] = - {x^-2[k^2xcos(kx) - ksin(kx)] - 2x^-3[cos(kx) + kxsin(kx)]} = - {[k^2xcos(kx) - ksin(kx)]/x^2 - [2cos(kx) + 2kxsin(kx)]/x^3} = - {[(kx)^2cos(kx) - kxsin(kx) - 2cos(kx) - 2kxsin(kx)]/x^3} = - {[(kx)^2cos(kx) - 2cos(kx) - 3kxsin(kx)]/x^3} = {[2cos(kx) + 3kxsin(kx) - (kx)^2cos(kx)]/x^3

d2y/dx^2 + 2/x.dy/dx + (k^2)y = {[2cos(kx) + 3kxsin(kx) - (kx)^2cos(kx)]/x^3} - {[2cos(kx) + 2kxsin(kx)]/x^3} + [k^2cos(kx)]/x = {[2cos(kx) + 3kxsin(kx) - (kx)^2cos(kx)]/x^3} - {[2cos(kx) + 2kxsin(kx)]/x^3} + [(kx)^2cos(kx)]/x^3 = {[2cos(kx) + 3kxsin(kx) - (kx)^2cos(kx) - 2cos(kx) - 2kxsin(kx) + (kx)^2cos(kx)]/x^3} = kxsin(kx)/x^3 = ksin(kx)/x^2

Where did I go wrong?
I thought it should be k^2xcos(kx) + ksin(kx), not minus
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BCHL85
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Ah no, [cos(kx) + kxsin(kx)]' = -ksin(kx) + k^2xcos(kx) + ksin(kx) = k^2.x.cos(kx).
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BCHL85
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#7
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hey.. you forgot k.x.sin(kx) = u.v where u = kx, v = sin(kx)
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BCHL85
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(Original post by Nima)
Sorry, I'm not sure what you're on about.
Hmm, ok let's derivative kxsin(kx).
kx = u, sin(kx) = v
-> u' = k, v' = kcos(kx)
So (uv)' = u'v + v'u = ksin(kx) + kcos(kx).kx
Is it alright now?
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BCHL85
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#9
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(Original post by Nima)
ahhh, of course, lol, I treated the x as a constant. oops.
Sometimes I have a mistake like that. :p:
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Nima
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#10
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(Original post by Bilal786)
Show that the function y = cos(kx) / x, where x =/ (does not equal) 0 and k is a non-zero constant, satisfies the differential equation (d^2)y/dx^2 + 2/x. dy/dx + (k^2)y = 0.
y = cos(kx)/x = x^-1.cos(kx)
dy/dx = (-x^-2)cos(kx) + (x^-1)(-ksin(kx)) = -[cos(kx)/x^2] - ksin(kx)/x
= -[cos(kx)/x^2] - kxsin(kx)/x^2 = [-cos(kx) - kxsin(kx)]/x^2 = -[cos(kx) + kxsin(kx)]/x^2 = -[cos(kx) + kxsin(kx)][x^-2]

d2y/dx^2 = - d/dx [cos(kx) + xksin(kx)][x^-2] = - {x^-2[k^2xcos(kx) + ksin(kx) - ksin(kx)] - 2x^-3[cos(kx) + kxsin(kx)]} = - {[k^2xcos(kx)]/x^2 - [2cos(kx) + 2kxsin(kx)]/x^3} = - {[(kx)^2cos(kx) - 2cos(kx) - 2kxsin(kx)]/x^3} = {[2cos(kx) + 2kxsin(kx) - (kx)^2cos(kx)]/x^3

d2y/dx^2 + 2/x.dy/dx + (k^2)y = {[2cos(kx) + 2kxsin(kx) - (kx)^2cos(kx)]/x^3} - {[2cos(kx) + 2kxsin(kx)]/x^3} + [k^2cos(kx)]/x = {[2cos(kx) + 2kxsin(kx) - (kx)^2cos(kx)]/x^3} - {[2cos(kx) + 2kxsin(kx)]/x^3} + [(kx)^2cos(kx)]/x^3 = {[2cos(kx) + 2kxsin(kx) - (kx)^2cos(kx) - 2cos(kx) - 2kxsin(kx) + (kx)^2cos(kx)]/x^3} = 0/x^3 = 0

QED BABY!
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BCHL85
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#11
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Woa, so hard-working, Nima... but I think it'd better use evariste's answer. Do you agree?
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BCHL85
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(Original post by Nima)
No....mines' so much more interesting.
Yeah ..it's good for doing practice about differentiation
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Bilal
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#13
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Cheers guys!
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