This is from S.L. Ross Differential Equation 3rd ed. I got confused where the form of the second solution coming from.
Consider the Bessel equation of order $p$ \begin{equation} x^2 \frac{d^2y}{dx^2} + x\frac{dy}{dx} + (x^2 - p^2)y = 0\tag{6.101} \end{equation}
Suppose this equation has solution of the form $\sum_{n = 0}^\infty c_nx^{n + r}$. Then we have the indicial equation is $r^2 - p^2 = 0$ and some other equations including recurrence relations for its coefficients namely $$[(r + 1)^2 - p^2]c_1= 0$$ and $$[(n + r)^2 - p^2]c_n + c_{n - 2} = 0,\quad n \geq 2.$$ For the roots $r = r_1 = p$ we got the first solution is Bessel equation of order $p$, $J_p$.
Now here where I got confused. For $r = r_2 = -p$ the equations above become $(-2p + 1)c_1= 0$ and $n(n - 2p)c_n + c_{n - 2} = 0$. The book says thi lead to the following cases for the solution
Case 1 if $2p \neq$ a positive integer, \begin{equation} y_2(x) = c_0x^{-p}\left(1 + \sum_{n = 1}^\infty \alpha_{2n} x^{2n}\right)\tag{6.130} \end{equation} Case 2 If $2p =$ an odd integer \begin{equation} y_2(x) = c_0x^{-p}\left(1 + \sum_{n = 1}^\infty \beta_{2n} x^{2n}\right) + c_{2p}x^{p}\left(1 + \sum_{n = 1}^\infty \gamma_{2n}x^{2n}\right)\tag{6.131} \end{equation} Case 3 If $2p =$ an even integer \begin{equation} y_2(x) = c_{2p}x^{p}\left(1 + \sum_{n = 1}^\infty \delta_{2n}x^{2n}\right)\tag{6.132} \end{equation} where $\alpha_{2n},\beta_{2n},\gamma_{2n},\delta_{2n}$ are definite constants and $c_0$ and $c_{2p}$ are arbitrary.
Here is my question, where did equations $6.132$ and $6.133$ coming from? I've try to find it the whole day but I just can't figure it out. Thanks in advance.