# How to solve Bessel?

• Apr 18th 2010, 08:51 AM
LooNiE
How to solve Bessel?
Ok so this is the problem that I am working on.

I have completed parts 1 and 2, which involve the changing of variables, but the problem I am having is finding one solution of the original DE.

This is the answer I got for part one is

http://img641.imageshack.us/img641/6483/42798846.jpg

The answer for part 2 is

http://img231.imageshack.us/img231/5672/70900969.jpg

I am sorry that I am using print screens, I am just not very good at using LateX and this was quicker.
• Apr 18th 2010, 08:52 PM
AllanCuz
Quote:

Originally Posted by LooNiE
Ok so this is the problem that I am working on.

http://img231.imageshack.us/img231/5672/70900969.jpg

I am sorry that I am using print screens, I am just not very good at using LateX and this was quicker.

I'm not going to go over all your simplification but assuming you are correct up to this point, you go astray in your last simplification. The equation becomes

$\displaystyle y^{\prime \prime} + y^{\prime} \frac{2}{x} + y(1 - \frac{3}{4x^2} )$

Note that x^2 is in the denominator. The way you had it is NOT a bessel equation. So now that we have this, what do we do?

Well, we know that the solutions to the following equation

$\displaystyle y^{\prime \prime } - \frac{2a-1}{x} y^{\prime} + (b^2 c^2 x^{2c-2} + \frac{a^2 - v^2c^2}{x^2} )y = 0$

are

$\displaystyle x^a J_v (bx^c)$

and

$\displaystyle x^a Y_v (bx^c)$

So let's compare this to your equation.

$\displaystyle y^{\prime \prime} + y^{\prime} \frac{2}{x} + (1 - \frac{3}{4x^2} )$

These are our equations

$\displaystyle 2a-1 = -2$

$\displaystyle b^2 c^2 = 1$

$\displaystyle 2c-2 = 0$

$\displaystyle a^2 - v^2 c^2 = - \frac {3}{4}$

We find the solutions to be

$\displaystyle a = -\frac{1}{2}$

$\displaystyle c=1$

$\displaystyle b=1$

$\displaystyle v = 1$

Therefore, the part of the general solution that they want is

$\displaystyle y=Bx^{ - \frac{1}{2} } J_v (x)$

Where

$\displaystyle x=2k \sqrt {s}$

$\displaystyle y=B (2k \sqrt {s} )^{ - \frac{1}{2} } J_1 (2k \sqrt {s})$

Of course K is a constant so we can let a new constant A absorb that value.

$\displaystyle y=A ( \sqrt {s} )^{ - \frac{1}{2} } J_1 (2k \sqrt {s})$

$\displaystyle y=A s^{ - \frac{1}{4} } J_1 (2k \sqrt {s})$

$\displaystyle \theta = \frac { A s^{ - \frac{1}{4} } J_1 (2k \sqrt {s}) } {\sqrt{x}}$

$\displaystyle \theta = \frac { A s^{ - \frac{1}{4} } J_1 (2k \sqrt {s}) } {\sqrt{2k}s^{1/4}}$

$\displaystyle \theta = Q s^{ - \frac{1}{2} } J_1 (2k \sqrt {s})$
• Apr 19th 2010, 01:06 AM
LooNiE
i think i have found my mistake (Happy). for part two, when i was finding the equation in terms of x and y, it was y(1-3/(4x^2)) which I can now multiply through by x^2 and it would be in bessel form. I'll now try and go over the steps that you showed me and hopefully I will get the right answer.
• Apr 19th 2010, 04:32 AM
LooNiE
ok im stuck again, and i still dont understand those substitutions you are making towards the start of your solution...

The function that I have now is

X^2(d2y/d2x)-2x(dy/dx)+(x^2 - 3/4)y=0

which I believe is of bessel type? Am i wrong?
• Apr 19th 2010, 07:15 AM
AllanCuz
Quote:

Originally Posted by LooNiE
ok im stuck again, and i still dont understand those substitutions you are making towards the start of your solution...

The function that I have now is

X^2(d2y/d2x)-2x(dy/dx)+(x^2 - 3/4)y=0

which I believe is of bessel type? Am i wrong?

This is a Bessel equation. But this will get you to the equation that I started off with.

Edit- We have worked through this, we are good now! Conclusion, prof missed a multiple :D