Solving d²Ψ/dx² + (2/x)dΨ/dx = λΨ

[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 through 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.