#### David Shulman

##### New member

- Joined
- Jul 29, 2020

- Messages
- 2

Hello,

I have a heterogeneous modified Bessel equation:

\(\displaystyle z'' +\frac{1}{r}z'-\frac{1}{v^2}z=G(r),\)

and I interesting in the solution of the function on extremum points,

\(\displaystyle z'(0)=0\)

so I solve the equation:

\(\displaystyle z''-v^2z=G(r),\)

I have for homogeneous solution:

\(\displaystyle z_h=c_1 exp(v^{-1} r)+c_2 exp(-v^{-1} r) \)

and for a Particular solution with Variation of Parameters:

\(\displaystyle z_p=exp(v^{-1} r)∫G(r)/(2v^{-1}) exp(-v^{-1} r)+exp(-v^{-1} r)∫G(r)/(2v^{-1} ) exp(v^{-1} r)\)

But how I can find the constant parameters in homogeneous solution? Probably I can not use my boundary condition, because I change my function...

I have a heterogeneous modified Bessel equation:

\(\displaystyle z'' +\frac{1}{r}z'-\frac{1}{v^2}z=G(r),\)

and I interesting in the solution of the function on extremum points,

\(\displaystyle z'(0)=0\)

so I solve the equation:

\(\displaystyle z''-v^2z=G(r),\)

I have for homogeneous solution:

\(\displaystyle z_h=c_1 exp(v^{-1} r)+c_2 exp(-v^{-1} r) \)

and for a Particular solution with Variation of Parameters:

\(\displaystyle z_p=exp(v^{-1} r)∫G(r)/(2v^{-1}) exp(-v^{-1} r)+exp(-v^{-1} r)∫G(r)/(2v^{-1} ) exp(v^{-1} r)\)

But how I can find the constant parameters in homogeneous solution? Probably I can not use my boundary condition, because I change my function...

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