# Hard Differentiation Question Watch

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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.

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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#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

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#3

(Original post by

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.

**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.

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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#4

look below,

(Original post by

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…

**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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#5

(Original post by

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[

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?

**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^3d2y/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?

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#8

(Original post by

Sorry, I'm not sure what you're on about.

**Nima**)Sorry, I'm not sure what you're on about.

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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#9

(Original post by

ahhh, of course, lol, I treated the x as a constant. oops.

**Nima**)ahhh, of course, lol, I treated the x as a constant. oops.

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#10

(Original post by

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.

**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.

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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#11

Woa, so hard-working, Nima... but I think it'd better use evariste's answer. Do you agree?

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#12

(Original post by

No....mines' so much more interesting.

**Nima**)No....mines' so much more interesting.

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