Hard Differentiation QuestionWatch

Announcements
This discussion is closed.
#1
I'm calling it hard because my teacher didn't know how to do it.

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.
0
14 years ago
#2
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
0
14 years ago
#3
(Original post by Bilal786)
I'm calling it hard because my teacher didn't know how to do it.

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
0
14 years ago
#4
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…
0
14 years ago
#5
(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
0
14 years ago
#6
Ah no, [cos(kx) + kxsin(kx)]' = -ksin(kx) + k^2xcos(kx) + ksin(kx) = k^2.x.cos(kx).
0
14 years ago
#7
hey.. you forgot k.x.sin(kx) = u.v where u = kx, v = sin(kx)
0
14 years ago
#8
(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?
0
14 years ago
#9
(Original post by Nima)
ahhh, of course, lol, I treated the x as a constant. oops.
Sometimes I have a mistake like that.
0
14 years ago
#10
(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!
0
14 years ago
#11
Woa, so hard-working, Nima... but I think it'd better use evariste's answer. Do you agree?
0
14 years ago
#12
(Original post by Nima)
No....mines' so much more interesting.
Yeah ..it's good for doing practice about differentiation
0
#13
Cheers guys!
0
X
new posts
Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

Thu, 24 Oct '19
• Cardiff University
Sat, 26 Oct '19
• Brunel University London
Sat, 26 Oct '19

Poll

Join the discussion

Yes (68)
23.45%
No (222)
76.55%