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.