Second Order Differential Equation

[danny271828]

I need to solve the equation

\frac{d^{2}}{dx^{2}}\Psi + \frac{2}{x}\frac{d}{dx}\Psi = \lambda\Psi

Can anyone help me get a start on this problem? I've been guessing at a few solutions with no results... I'm not asking anyone to solve the problem... just a few hints on starting... maybe regarding the form of the solution... Thanks

 

[Kummer]

I need to solve the equation

\frac{d^{2}}{dx^{2}}\Psi + \frac{2}{x}\frac{d}{dx}\Psi = \lambda\Psi

Can anyone help me get a start on this problem? I've been guessing at a few solutions with no results... I'm not asking anyone to solve the problem... just a few hints on starting... maybe regarding the form of the solution... Thanks

Multiply throught by x^2 to get,
x^2 \Psi '' + 2x\Psi ' - \lambda x^2 \Psi = 0
Now I believe this can be converted to Bessel's equations.
The solution is in terms of Bessel functions (possibly using the Bessel function of the 3rd kind, i.e. Hankel function).

 

[HallsofIvy]

If it is not a form of Bessel's equation, use Frobenius' method.

 

[chaoseverlasting]

Dont know if this would work, but you could substitute p=dy/dx, and convert this into a linear DE in p. Once you get a solution for p, replace p by dy/dx and that's another linear DE to solve.

 

[HallsofIvy]

Dont know if this would work, but you could substitute p=dy/dx, and convert this into a linear DE in p. Once you get a solution for p, replace p by dy/dx and that's another linear DE to solve.

But there is no "y" in the problem! Presumably you meant \Psi but then, since \Psi itself appears in the equation, that substitution won't work.

 

[chaoseverlasting]

But there is no "y" in the problem! Presumably you meant \Psi but then, since \Psi itself appears in the equation, that substitution won't work.

Yeah. Sorry... Didnt realize that.