Bessel equation

**panda** · June 13th 2008, 04:04 AM

the question says 'use the result to solve bessel's equation of index n=1/2'.
i wasn't sure what it means by index n=1/2?

**CaptainBlack** · June 13th 2008, 05:18 AM

Originally Posted by panda

the question says 'use the result to solve bessel's equation of index n=1/2'.
i wasn't sure what it means by index n=1/2?

Compute $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ in terms of $x$ and $\nu$ using the suggested $y(x)=x^{-1/2}\nu(x)$ . It is labourious but it does give the quoted equation.

Now if we put $n=1/2$ in the final equation it reduces to:

$<br /> \frac{d^2\nu}{dx^2}+\nu=0<br />$

which has general solution $\nu(x)=A\cos(x)+B\sin(x)$ , so the solution of the Bessel equation with $n=1/2$ (that is of order $n=1/2$ ) is:

$<br /> y(x)=x^{-1/2}[A\cos(x)+B\sin(x)]<br />$

RonL

Thread: Bessel equation

LinkBack

Thread Tools

Display

Bessel equation

Similar Math Help Forum Discussions

Bessel equation

Bessel's equation

Bessel's equation

Bessel Equation

Bessel differencial equation

Search Tags