Transforming a DE into Bessel's Equation

[vertices]

I am trying to understand why the General solution to this DE:

x\frac{d^2y}{dx^2} +y=0 (1)

is:

y(x)=x^{\frac{1}{2}}\left(AJ_1(2x^{\frac{1}{2}})+BY_1(2x^{\frac{1}{2}})\right) (*)

This is part of a physics problem - the solutions (as in 'answers') don't offer any explanation - they simply say:

y'' + \frac{1-2a}{x}y' + \left((bcx^{c-1})^2+ \frac{a^2 - p^2 c^2}{x^2}\right)y=0 (2)

where

y=x^aZ_p(bx^c) (3)

Z_p satisfies the Bessel Eqn of order p.

If we equate the coefficients of (2) with the original DE (1), this gives us the solution (*).

I really can't see where (2) has come from. I've tried substituting in (3) into the original DE and get a horribly complicated expression which doesn't look anything like (2) - also I can't understand why 'p', the order of the Bessel's Equation, enters the DE (2).

Any help would be much appreciated.

thanks.

 

[LCKurtz]

I would suggest trying to solve your DE directly. x = 0 is a regular singular point, so you would look for a solution of the form

y = \sum_0^\infty a_nx^{n+r}

Plug that in the equation using usual series methods. You should get an indicial equation of (r)(r-1)= 0. The roots differ by an integer so you can get one solution using r = 1. If you work it out you should get

a_n = \frac{(-1)^n a_0}{(n+1)!n!}

giving a solution

y(x) = \sum_0^\infty \frac{(-1)^na_0}{(n+1)!n!}\ x^{n+1}

Remembering that

J_1(x)= \sum_0^\infty \frac{(-1)^n}{n!\Gamma(n+2)}\left(\frac x 2\right)^{2n+1}

Using this you should be able to calculate

\sqrt xJ_1(2\sqrt x)

and get the series for y(x) above. You will have to review how to get the second solution when the indicial roots differ by an integer, but this should get you started.

 

[vertices]

Thanks LCKurtz -

Can I ask a stupid question though (my maths knowledge is very superficial): if we know that J₁(x) is a solution to a DE, isn't it always the case that Y₁(x) must also be part of the general solution (as it is linearly independent to to J₁)?

Also can you offer any insight on why the original DE can be transformed(?) into this DE:

y'' + \frac{1-2a}{x}y' + \left((bcx^{c-1})^2+ \frac{a^2 - p^2 c^2}{x^2}\right)y=0

where

y=x^aJ_p(bx^c)

 

[jackmell]

I am trying to understand why the General solution to this DE:

x\frac{d^2y}{dx^2} +y=0 (1)

is:

y(x)=x^{\frac{1}{2}}\left(AJ_1(2x^{\frac{1}{2}})+BY_1(2x^{\frac{1}{2}})\right) (*)

Hi. You know about Mathworld? Google it and then do a Bessel equation search and scroll down to the Bowman forumulation of Bessel's equation. The one with all the parameters. Now substitute your equation into that one. I get \alpha=1/2, n=1, \gamma=1/2, \beta=\pm 2 then substitute those into the solution given to arrive at

y(x)=x^{\frac{1}{2}}\left(AJ_1(2x^{\frac{1}{2}})+BY_1(2x^{\frac{1}{2}})\right)

I'm not too sure about the \pm \beta part though.