Solving Bessel's Equation: Homework

[TheFerruccio]

Homework Statement

Find a general solution in terms of Bessel functions. (Use the indicated transformations and show the details)

Homework Equations

$$x^2y''+5xy'+(x^2-12)y = 0$$
$$y = \frac{u}{x^2}$$

The Attempt at a Solution

I know that the answer needs to be in the form of Bessel functions of the first and second kind, depending on what my roots are.

What is throwing me off is the substitution. The book never explained the nature of substitution for this type of problem. I just literally substituted in that variable like they told me, but I have an inkling that this is extremely wrong, and is going in the completely wrong direction:

$$x^2(6ux^{-4})+5x(-2ux^{-3})+(x^2-12)ux^{-2} = 0$$
$$-16ux^{-2} + u = 0$$

After this, I reached a dead end. I do not know how to relate this to finding my roots of r from the indicial equation. I have a sense that I need to use the series method first, then somewhere in the middle of the problem, do the substitution.

 

[vela]

You didn't calculate the derivatives correctly. Remember that u is a function of x, so when you calculate y'' and y', you have to use the quotient rule.

 

[TheFerruccio]

You didn't calculate the derivatives correctly. Remember that u is a function of x, so when you calculate y'' and y', you have to use the quotient rule.

Ohhh, I thought $$u$$ was just a constant.

 

[TheFerruccio]

EDIT: I made two mistakes here. First: I forgot to convert the substitution back. Second: I forgot that the values of $$\nu$$ differ by an integer, which necessitates using the Bessel function of the second kind for the other solution.


Here's an attempt to solve it. I believe that, since $$u$$ is a function of $$x$$, then I am to use the product rule to find all the associated derivatives, then plug into the original equation in order to come up with Bessel's differential equation.

$$y = \frac{u}{x^2}$$
$$-2ux^{-3}+u'x^{-2}$$
$$6ux^{-4}-2u'x^{-3}+u''x^{-2}-2u'x^{-3}$$


Plugging in...

$$x^{2}(6x^{-4}u-2x^{-3}u'+x^{-2}u''-2x^{-3}u')+5x(-2x^{-3}u+x^{-2}u')+x^2(x^{-2}u)-12(x^{-2}u) = 0$$

This leaves me with...

$$u''+\frac{u'}{x}+(\frac{-16}{x^2}+1) = 0$$

multiplying by $$x^2$$...

$$x^2u''+xu' + (x^2-16)u = 0$$

so

$$\nu=\pm4$$

Using the Bessel function of the first kind of order $$\nu$$ I get the first solution...

$$u_1(x) = y_1(x)x^2 = c_1x^4\sum_{m=0}^{\infty}\frac{(-1)^mx^{2m}}{2^{2m+4}m!\Gamma(m+5)}$$

The second solution uses the Bessel function of the second kind.

$$u_2(x) = \frac{1}{\sin{4\pi}}[J_{4}(x)\cos{4\pi}-J_{-4}(x)]$$

 

[vela]

Two problems: First, the solutions to the differential equation for u(x), not y(x), are the Bessel functions. Second, when ν is an integer, J_ν and J_-ν aren't linearly independent, so you still need to find a second independent solution for u(x).

 

[TheFerruccio]

Disregard the previous post I made. I have the problem of solving my confusion while typing the post, which results in countless edits and, consequently, likely confusion for the helpers here who are contributing their time. Sorry.

 

[vela]

The second solution uses the Bessel function of the second kind.

$$u_2(x) = \frac{1}{\sin{4\pi}}[J_{4}(x)\cos{4\pi}-J_{-4}(x)]$$

This doesn't work either because sin 4π=0 and J₄=J_-4. You need to write it as a limit. In any case, you're probably just expected to express u(x) as a linear combination of J₄(x) and Y₄(x). You don't need to write out what the functions actually equal.

 

[TheFerruccio]

This doesn't work either because sin 4π=0 and J₄=J_-4. You need to write it as a limit. In any case, you're probably just expected to express u(x) as a linear combination of J₄(x) and Y₄(x). You don't need to write out what the functions actually equal.

Again, right (read the first part of my post, I stuck an edit there yesterday, I realized both mistakes and later said to disregard that post, lol).

Anyway, here's the solution I got, for the sake of closure (not the mathematical closure):

$$y(x) = c_1\frac{J_4(x)}{x^2}+c_2\frac{Y_4(x)}{x^2}$$ (edited for typo)

where
$$J_4(x)=x^4\sum_{m=0}^{\infty}\frac{(-1)^mx^{2m}}{2^{2m+4}m!(M+4)!}$$

and

$$Y_4(x) = \frac{2}{\pi}x^4J_4(x)(\ln(\frac{x}{2})+ \gamma)+\frac{x^4}{\pi}\sum_{m=0}^{\infty}\frac{(-1)^{m-1}(\sum_{j=1}^{\infty}\frac{1}{j}+\sum_{j=1}^{m+4}\frac{1}{j})}{2^{2m+4}m!(m+4)!}x^{2m}-\frac{x^{-4}}{\pi}\sum_{m=0}^{3}\frac{(3-m)!}{2^{2m-4}m!}x^{2m}$$

where

$$\gamma = \lim_{n \to 0}(\sum_{k=1}^{n}\frac{1}{k} - \ln(n))$$ AKA the Euler–Mascheroni constant, which is $$\approx .577$$
Thanks for the help!