# Help with special function differential equation

this is my first time to use this site. Please let me know if the equations are unreadable, latex isn't my first language.

We've been covering Legendre, Bessel, and Confluent Hypergeometric Functions as special functions frequently observed in solving differential and integral equations. This chapter is difficult to understand.

I seem to be hitting a wall and/or overthinking the following:

Suppose a linear second order differential equation has the following solution:

$$x^{\alpha }J_{\pm m}(\beta x^{\gamma })$$

What differential equation might this be? Using this, give the general solution of: $$y''+x^{2}y=0$$

Any help is appreciated.

If you have a DEQ of the form:

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

It has the solution:

$$y = x^aZ_p(bx^c).$$

where $Z$ stands for $J$ or $N$ or any linear combination of them and $a,b,c,p$ are constants. All you do is match up the and solve for the constants from your given DEQ and then you have the solution $y$.

You should be able to find this in the NIST DLMF or Math World.

When you write the solution, you should verify it as you've likely learned in class.

Also, just to be complete, this DEQ can use the Parabolic Cylinder Function as a solution.

• Great to know, and nice links provided +1 – Namaste Apr 28 '13 at 0:09
• It would be a great resource to have a resource link/tab available on Math.SE, listed according to domains, etc. Or, along with a tag's FAQ – Namaste Apr 28 '13 at 0:15

Look here. The basic idea is: you know that $g(t)=J_{\pm n}(t)$ satisfy Bessel equation, then one should look at what happens to this equation if we change the independent variable, $t=\beta x^{\gamma}$, and simultaneously replace the function $g(t)$ by $x^{-\alpha}y(x)=\left(t/\beta\right)^{-\alpha/\gamma}y\left(\left(t/\beta\right)^{1/\gamma}\right)$.

Concerning the specific equation in the end, compare it with (6): then $2\alpha-1=0$, $\alpha^2-n^2\gamma^2=0$, $2\gamma-2=2$, $\beta^2\gamma^2=1$.