Differential equation reducible to Bessel's Equation

1. Jan 18, 2008

joel19

How do I reduce an equation to Bessel's equation and find a general solution to it:

For example how do I solve this:

x^2y" + xy' + (4x^4 - 1/4)y = 0 (set x^2 = z)

2. Jan 18, 2008

HallsofIvy

Staff Emeritus
Okay, apparently you have been told to "set z= x2". Have you tried that at all? Do you know how to use the chain rule to convert d2y/dx2 to d2y/dz2?

To start you off, dy/dx= (dy/dz)(dz/dx)= (2x)(dy/dz).

Then d^2y/dx^2= d(2x dy/dz)/dx= 2 dy/dz+ d(dy/dx)/dx= 2 dy/dz +(d^2y/dz^2)(2x).

3. Nov 2, 2008

srknkcyrk

Hi all,

How do i reduce the eqn. below to Bessel's eqn. (how can i use the transformation when i have y, x, and u)

4x^2y'' - 20xy' + (4x^2 + 35)y = 0 (y = (x^3).u)

4. Nov 2, 2008

HallsofIvy

Staff Emeritus
Do you mean to set y= x3u(x) where u is some unknown function? Then just DO it!

If y= x3u, then y'= 3x2u+ x3u' and y"= 6xu+ 6x2u'+ x3u". Replace y, y' and y" in the equation with those and you get another differential equation for u rather than y.

5. Nov 2, 2008

srknkcyrk

I was not sure about u is a function of x, should that be so? I am asking this is because i do not understand (unless u is a func. of x) why we use a third variable for reducing.

I will try what you suggested, and thank you for your help.

6. Nov 2, 2008

HallsofIvy

Staff Emeritus
If u is NOT a function of x, what is it? A constant? If that were so, you would be asking how to write an equation in x and y, a function of x, in terms of x only, and that cannot be done.

7. Nov 2, 2008

srknkcyrk

Yes, you are right. I just could not think simple, (may be it can be written as u(x) instead of u, for being more clear, but u is commonly used as a function of x, i should have remembered that),anyway, i reduced the ode to Bessel's eqn. Thanks for your help again.

8. Jul 27, 2009

xerxex_dan

hey pips can you help me solve this one:

xy'' - y' + y = 0....in terms of bessel functions

9. Jul 30, 2009

matematikawan

Try this substitution:

$$X=2\sqrt{x}$$

$$Y=\frac{y}{x}$$

10. Sep 23, 2010

abhishekgoyal

Hey could you help me solve
x^2*y''+2x*y'+x^2*y=0