Removing a Singularity for a Second Order ODE

[DrChops]

I was given the following equation to solve:

x^2*y'' + x*y' + k^2*x^2*y = 0
B.C. y'(0)=0, y(1)=0

where k is just some constant.

I am having a hard time removing the singularity created by the boundary condition at y' and not aware of a method how. Any advice would be greatly appreciated.

 

[haruspex]

For x nonzero, you can obviously simplify the equation. When x is zero, there simply is no information from the DE. So any solution of the simplified equation is a valid solution for the original.
Please show your working so far.

 

[LCKurtz]

Not sure what you mean by removing the singularity. You have a regular singular point at $x=0$. Have you looked for a solution of the form$$
y=\sum_{n=0}^\infty x^{n+r}\hbox{?}$$

 

[JJacquelin]

Hi !

it is a Bessel ODE:
y = c1*J0(k*x) +c2*Y0(k*x)
The condition y'(0)=0 implies c2=0, hence y(x) = c1*J0(k*x)
J(k)=0 only in case of some particular values of k.
So, generally, the condition y(1)=0 implies c1=0, hence y(x)=0 in general.
But, y(x)=c1*J0(k*x) in case of some particular values of k.

 

[DrChops]

Thanks for all the help so far.
I'm just trying to express the Bessel function in a way that satisfies the boundary conditions. So far I have written it out to:
1- 1/4*k^2*x^2 + (.25*k^2*x^2)^2/4 - (1/4*k^2*x^2)^3/
36 + (.25*x^2*k^2)^4/24^2

Using equation 78 off this link http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
My problem is that I can't get it to meet the boundary conditions. If I could take out the 1 and set "k" equal to about 5.3, it fits... but I think that's cheating.

 

[haruspex]

Does this help:
Let r_i be the i^th +ve root of J1(x). So it also satisfies J0'(x) = 0.
Consider Y(x) = Ʃ a_iJ0(r_ix)
This satisfies Y'(1) = 0.
If we can find a_i s.t. Ʃa_ir_i² = k²Ʃ a_i then I believe Y satisfies the original equation, and Y'(0) = Ʃ a_ir_iJ0'(0) = 0