How do you solve for f in the differential equation (x^2-a^2)f'+xf=0?

[Helios]

This one looks easy but I don't know.

( x^{2} - a^{2} )f' + xf = 0

 

[DeadWolfe]

Isn't that separable?

 

[Matthew Rodman]

Rearrange to get

\frac{f^{\prime}}{f} = - \frac{x}{x^2 - a^2}

now integrate to get

\ln{f} = -\frac{1}{2} \ln{(x^2 - a^2)} + C