Differential equation reducible to Bessel's Equation

[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)

 

[HallsofIvy]

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).

 

[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)

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

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.

 

[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.

 

[xerxex_dan]

hey pips can you help me solve this one:

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

 

[matematikawan]

Try this substitution:

X=2\sqrt{x}

Y=\frac{y}{x}

 

[abhishekgoyal]

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