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,
$\displaystyle \frac{D^2 \theta}{dS^2} + \frac{2d \theta}{sdS} + \frac{k^2 \theta}{s}=0$
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.
$\displaystyle \frac{D^2 \theta}{dS^2} + \frac{k^2 \theta}{s}= - \frac{2d \theta}{sdS} $
$\displaystyle \frac{s D^2}{dS} + k^2 dS= - \frac{2d \theta}{ \theta } $
Noting that
$\displaystyle x=2k \sqrt{s}$
$\displaystyle s = \frac{x^2}{4k^2}$
Going with
$\displaystyle \frac {dS}{dX}= \frac {2x}{4k^2}$
$\displaystyle dS= dX \frac {2x}{4k^2}$
Sub this into our original equation to get
$\displaystyle \frac{s D^2}{ \frac{2xdX}{4k^2}} + k^2 ( \frac{2xdX}{4k^2} )= - \frac{2d \theta}{ \theta } $
$\displaystyle \frac{4k^2 s D^2}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta } $
Now we sub in
$\displaystyle s = \frac{x^2}{4k^2}$
To get
$\displaystyle \frac{4k^2 D^2 \frac{x^2}{4k^2}}{ 2xdX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta } $
$\displaystyle \frac{D^2 x}{ 2dX} + ( \frac{xdX}{2} )= - \frac{2d \theta}{ \theta } $
$\displaystyle \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!