A little confusion I have got, In this question (in the middle), $$\chi''(r) + \frac{d - 1 + 2\beta}{r}\chi'(r) + \left(k^2 + \frac{\beta(\beta + d - 2)}{r^2}\right)\chi(r) = 0.$$ Form this line he inserted $\beta = 1 - \frac{d}{2}$, and got, $$\chi''(r) + \frac{1}{r} \chi'(r) + \left(k^2 - \frac{(1-\frac{d}{2})^2}{r^2}\right)\chi(r) = 0.$$
I have inserted the same value but it isn't matching :-(. the problem to match the factor $\frac{(1-\frac{d}{2})^2}{r^2}$
Am I missing something? or there is a problem in solving?
Ideally, in the future, you should simply leave a comment on the answer in question, asking for clarification.
The reason this works is because $$\begin{align}\beta+d-2 &= 1-\frac{d}2+d-2\\ &= \frac d2-1\\ &=-\left(1-\frac d2\right).\end{align}$$ I suspect that you simply made an arithmetic error.