Let: :$\displaystyle y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$ such that: :$a_0 \ne 0$ Differentating {{WRT|Differentiation}} $x$: :$\displaystyle \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$ :$\displaystyle \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$ Substituting in the original equation: {{begin-eqn}} {{eqn | l = \paren {1 - x^2} \ddot y - 2 x \dot y + p \paren {p + 1} y       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {1 - x^2} \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2} - x^2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}       | o =        | c =  }} {{eqn | l = {} - 2 x \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}       | r = 0       | c =  }} {{end-eqn}} The summations are dependent upon $n$ and not $x$. Therefore it is a valid operation to multiply the $x$'s into the summations, thus: {{begin-eqn}} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}       | o =        | c =  }} {{eqn | l = {} - \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n} - 2 \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n} + p \paren {p + 1} \sum_{n \mathop = 0}^\infty a_n x^{k - n}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}       | o =        | c =  }} {{eqn | l = {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n - 1} - 2 \paren {k - n} }       | r = 0       | c =  }} {{end-eqn}} Increasing the summation variable $n$ to $n + 2$: {{begin-eqn}} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}       | o =        | c =  }} {{eqn | l = {} + \sum_{n \mathop = 0}^\infty a_n x^{k - n} \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }       | r = 0       | c =  }} {{end-eqn}} Taking the first 2 terms of the second summation out: {{begin-eqn}} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 2}^\infty a_{n - 2} \paren {k - n + 2} \paren {k - n + 1} x^{k - n}       | o =        | c =  }} {{eqn | l = {} + a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }       | o =        | c =  }} {{eqn | l = {} + \sum_{n \mathop = 2}^\infty a_{n - 2} x^{k - n} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } + a_1 x^{k - 1} \paren {\paren {p + 1} - k \paren {k - 1} }       | o =        | c =  }} {{eqn | l = {} + \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }       | r = 0       | c =  }} {{end-eqn}} Equating each term to $0$: {{begin-eqn}} {{eqn | n = 1       | l = a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} }       | r = 0       | c =  }} {{eqn | n = 2       | l = a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} }       | r = 0       | c =  }} {{eqn | n = 3       | l = \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} +  a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } }       | r = 0       | c =  }} {{end-eqn}} Take equation $(1)$: :$a_0 x^k \paren {p \paren {p + 1} - k \paren {k + 1} } = 0$ It is assumed that $a_0 \ne 0$ and $x^k$ can never be zero for any value of $k$. Thus: {{begin-eqn}} {{eqn | l = p \paren {p + 1} - k \paren {k + 1}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = p^2 - k^2 + p - k       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {p - k} \paren {p + k} + \paren {p - k}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {p - k} \paren {p + k + 1}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = p - k       | r = 0       | c =  }} {{eqn | lo= \lor       | l = p + k + 1       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = k       | r = p       | c =  }} {{eqn | lo= \lor       | l = k       | r = p - 1       | c =  }} {{end-eqn}} Take equation $(2)$: :$a_1 x^{k - 1} \paren {p \paren {p + 1} - k \paren {k - 1} } = 0$ As before, it is assumed that $x^{k - 1}$ can never be zero for any value of $k$. Thus: {{begin-eqn}} {{eqn | l = a_1 \paren {p \paren {p + 1} - k \paren {k - 1} }       | r = 0       | c =  }} {{eqn | l = a_1 k       | r = 0       | c = substituting the value of $k$ from $(1)$ }} {{eqn | l = a_1       | r = 0       | c = as $2 k \ne 0$ }} {{end-eqn}} Take equation $(3)$: :$\displaystyle \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$ As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$. Thus: {{begin-eqn}} {{eqn | l = 0       | r = a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = a_n       | r = -\frac {\paren {k - n + 2} \paren {k - n + 1} } {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } a_{n - 2}       | c =  }} {{end-eqn}} Since Legendre's differential equation is a second order ODE, it has two independent solutions. === Solution 1 (for $k = p$) === {{begin-eqn}} {{eqn | l = y       | r = x^p \sum_{n \mathop = 0}^\infty a_n x^{-n}       | c =  }} {{eqn | r = x^k \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_n       | r = -\frac {\paren {p - n + 2} \paren {p - n + 1} } {p \paren {p + 1} - \paren {p - n} \paren {p - n + 1} } a_{n - 2}       | c = from equation $(4)$ }} {{eqn | r = -\frac {\paren {p - n + 2} \paren {p - n + 1} } {n \paren {2 p - n + 1} } a_{n - 2}       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = a_2       | r = -\frac {p \paren {p - 1} } {2 \paren {2 p - 1} } a_0       | c = for $n = 2$ }} {{eqn | l = a_4       | r = -\frac {\paren {p - 2} \paren {p - 3} } {4 \paren {2 p - 3} } a_2        | c = for $n = 4$ }} {{eqn | r = \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } a_0       | c =  }} {{eqn | l = a_6       | r = -\frac {\paren {p - 4} \paren {p - 5} } {6 \paren {2 p - 5} } a_4       | c = for $n = 6$ }} {{eqn | r = -\frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \paren {p - 4} \paren {p - 5} } {2 \cdot 4 \cdot 6 \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} } a_0       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a_{2 n}       | r = \paren {-1}^n \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} \dotsm \paren {p - 2 n - 1} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p - 1} \paren {2 p - 3} \paren {2 p - 5} \dotsm \paren {2 p - 2 n + 1} } a_0       | c =  }} {{end-eqn}} Also: :$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$ Substituting for $a_n$: {{begin-eqn}} {{eqn | l = y       | r = a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\right.       | c =  }} {{eqn | o =       | ro= +       | r = \dotsb       | c =  }} {{eqn | o =       | ro= +       | r = \left.{\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\right)       | c =  }} {{end-eqn}} {{qed|lemma}} === Solution 2 (for $k = p - 1$) === {{begin-eqn}} {{eqn | l = y       | r = x^{-p - 1} \sum_{n \mathop = 0}^\infty a_n x^{-n}       | c =  }} {{eqn | r = x^{-k - 1} \paren {a_0 + a_1 x^{-1} + a_2 x^{-2} + a_3 x^{-3} + a_4 x^{-4} + \dotsb}       | c =  }} {{end-eqn}} {{questionable|Should it be $x^{-k - 1}$ in the above, as $k {{=}} p - 1$?}} {{begin-eqn}} {{eqn | l = a_n       | r = -\frac {\paren {-p - 1 - n + 2} \paren {-p - 1 - n + 1} } {p \paren {p + 1} - \paren {- p - 1 - n} \paren {-p - n} } a_{n - 2}       | c = from equation $(4)$ }} {{eqn | r = \frac {\paren {p + n - 1} \paren {p + n} } {n \paren {2 p + n + 1} } a_{n - 2}       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = a_2       | r = \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } a_0       | c = for $n = 2$ }} {{eqn | l = a_4       | r = \frac {\paren {p + 3} \paren {p + 4} } {4 \paren {2 p + 5} } a_2       | c = for $n = 4$ }} {{eqn | r = \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } a_0       | c =  }} {{eqn | l = a_6       | r = \frac {\paren {p + 5} \paren {p + 6} } {6 \paren {2 p + 7} } a_4       | c = for $n = 6$ }} {{eqn | r = \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} \paren {p + 5} \paren {p + 6} } {2 \cdot 4 \cdot 6 \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} } a_0       | c =  }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = a_{2 n}       | r = \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \paren {2 p + 7} \dotsm \paren {2 p + 2 n + 1} } a_0       | c =  }} {{end-eqn}} Also: :$a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$ Substituting for $a_n$: {{begin-eqn}} {{eqn | l = y       | r = a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\right.       | c =  }} {{eqn | o =       | ro= +       | r = \dotsb       | c =  }} {{eqn | o =       | ro= +       | r = \left.{\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\right)       | c =  }} {{end-eqn}} {{qed|lemma}} In summary, the two particular solutions are: {{begin-eqn}} {{eqn | l = y       | r = a_0 x^k \left({1 + \frac {-p \paren {p - 1} } {2 \paren {2 p - 1} } x^{-2} + \frac {p \paren {p - 1} \paren {p - 2} \paren {p - 3} } {2 \cdot 4 \paren {2 p - 1} \paren {2 p - 3} } x^{-4} }\right.       | c =  }} {{eqn | o =       | ro= +       | r = \dotsb       | c =  }} {{eqn | o =       | ro= +       | r = \left.{\paren {-1}^n \frac {-p \paren {p - 1} \paren {p - 2} \dotsm \paren {p - 2 n + 1} } {\paren {2 \cdot 4 \cdot 6 \dots 2 n} \paren {2 p - 1} \paren {2 p - 3} \dotsm \paren {p - 2 n + 1} } x^{-2 n} + \dotsb}\right)       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = y       | r = a_0 x^{-k - 1} \left({1 + \frac {\paren {p + 1} \paren {p + 2} } {2 \paren {2 p + 3} } x^{-2} + \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \paren {p + 4} } {2 \cdot 4 \paren {2 p + 3} \paren {2 p + 5} } x^{-4} }\right.       | c =  }} {{eqn | o =       | ro= +       | r = \dotsb       | c =  }} {{eqn | o =       | ro= +       | r = \left.{\paren {-1}^n \frac {\paren {p + 1} \paren {p + 2} \paren {p + 3} \dotsm \paren {p + 2 n} } {\paren {2 \cdot 4 \cdot 6 \dotsm 2 n} \paren {2 p + 3} \paren {2 p + 5} \dotsm \paren {p - 2 n} } x^{-2 n} + \dotsb}\right)       | c =  }} {{end-eqn}} {{improve|Express in summation form}} {{qed}}