Why Must B Approach Zero in Cauchy-Euler ODE Solutions at x=0?

[Andreol263]

Why these ODEs when applied some boundary conditions, like x = 0, their solution of the form Ax^k + Bx^(-k), B WILL have to go to zero?Like some problems which involve spherical harmonics...

 

[jambaugh]

The negative exponent of the B coefficient term implies divergent solution at x=0. If you are considering solutions with physical interpretation such as electron wave functions then the physical application is contradicted by the divergence at infinity. Those solutions don't apply and you set B = 0.

 

[Andreol263]

Yeah i understand why when x->infinity the solution is inconsistent, cause it's blow up, but i simply don't get it WHY x=0 implies in a divergent solution and B has to be 0, there's a more deep explanation?

 

[jambaugh]

$B x^{-k}=\frac{B}{x^k} = \frac{B}{0}$ when $x=0$.

 

[Andreol263]

Oh damn, I'm stupid , i was thinking in that, but it appears so simple that i forget this option because some texts are very confusing , and thank you so much for killing that existential question for me :D...