Correcting Solutions for Euler's Equation with Kronecker Delta Function

[Precursor]

How do I solve the following Euler's equation:

r^2 B_n'' + r B_n' - n^2 B_n = 3 \delta_{n1} r^2

Such that the solution is:

B_n(r) = \beta_n r^n + \delta_{n1}r^2, \forall n \ge 1

where βn is a free coefficient, δ is the Kronecker delta function, and the solutions unbounded at r=0 are discarded.

 

[Dick]

How do I solve the following Euler's equation:

r^2 B_n'' + r B_n' - n^2 B_n = 3 \delta_{n1} r^2

Such that the solution is:

B_n(r) = \beta_n r^n + \delta_{n1}r^2, \forall n \ge 1

where \beta_{n} is a free coefficient, \delta is the Kronecker delta function, and the solutions unbounded at r = 0 are discarded.

You try to solve the differential equation for different values of n. n=1 is obviously different from all other n. The trick for an equation in this form is to use a trial function of the form $B_n(r)=Cr^k$ and solve for k.

 

[Precursor]

Is the \delta_{n1}r^2 obtained in the solution by linearity? And why is the coefficient '3' not in front?

 

[Dick]

Is the \delta_{n1}r^2 obtained in the solution by linearity? And why is the coefficient '3' not in front?

I'm not sure what you are asking. The case n=1 is different from the other values of n because then the right side is 3r^2. If n is not 1 then the right side is 0. That's what the Kronecker delta does. They are two different cases. Solve them separately.

 

[Precursor]

I've followed your steps, but when I'm solving the case for n = 1, I get B_{1} = \beta_{1}r + r^{2}. Shouldn't it only be r^{2}?

 

[Dick]

I've followed your steps, but when I'm solving the case for n = 1, I get B_{1} = \beta_{1}r + r^{2}. Shouldn't it only be r^{2}?

Yes, it should. You have to put $\beta_{1}=0$ in that case. The problem statement is sloppy.