Help substituting !

• Apr 15th 2010, 07:04 AM
LooNiE
Help substituting !
I have

(D^2theta/ds^2)+((2/s)dtheta/ds)+((k^2)/s)theta=0

I am asked to make the subtitution x=2k*sqrt(s) in order to get the above in terms of theta and x.

How do i do this?
• Apr 15th 2010, 10:25 AM
AllanCuz
Quote:

Originally Posted by LooNiE
I have

(D^2theta/ds^2)+((2/s)dtheta/ds)+((k^2)/s)theta=0

I am asked to make the subtitution x=2k*sqrt(s) in order to get the above in terms of theta and x.

How do i do this?

I'm assuming you mean,

$\frac{D^2 \theta}{dS^2} + \frac{2d \theta}{sdS} + \frac{k^2 \theta}{s}=0$

edit- Mulligen, I think i violated some rules here...1 momento

I'm not entirely sure off the bat where to go with this. But let us try some manipulation.

$\frac{D^2 \theta}{dS^2} + \frac{k^2 \theta}{s}= - \frac{2d \theta}{sdS}$

$\frac{s D^2}{dS} + k^2 dS= - \frac{2d \theta}{ \theta }$

Noting that

$x=2k \sqrt{s}$

$s = \frac{x^2}{4k^2}$

Going with

$\frac {dS}{dX}= \frac {2x}{4k^2}$

$dS= dX \frac {2x}{4k^2}$

Sub this into our original equation to get

$\frac{s D^2}{ \frac{2xdX}{4k^2}} + k^2 ( \frac{2xdX}{4k^2} )= - \frac{2d \theta}{ \theta }$

$\frac{4k^2 s D^2}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }$

Now we sub in

$s = \frac{x^2}{4k^2}$

To get

$\frac{4k^2 D^2 \frac{x^2}{4k^2}}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }$

$\frac{D^2 x}{ 2dX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta }$

$\frac{D^2 x}{ dX} + xdX= - \frac{d \theta}{ \theta }$

And since I had no idea where to begin, I have run out of manipulations that I can think of. Hopefully this will prompt some of your own!
• Apr 15th 2010, 10:46 AM
LooNiE
Yes, that's what I meant.
• Apr 17th 2010, 06:24 PM
AllanCuz
I'd like to see an answer to this if you guys wouldn't mind (Bow) I'll work on it more when i decide i need more of a break from exam studying.