Solution to Legendre's Differential Equation

Theorem
The solution of Legendre's differential equation:

can be obtained by Power Series Solution Method.

Proof
Let:
 * $\ds y = \sum_{n \mathop = 0}^\infty a_n x^{k - n}$

such that:
 * $a_0 \ne 0$

Differentating $x$:


 * $\ds \dot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} x^{k - n - 1}$


 * $\ds \ddot y = \sum_{n \mathop = 0}^\infty a_n \paren {k - n} \paren {k - n - 1} x^{k - n - 2}$

Substituting in the original equation:

The summations are dependent upon $n$ and not $x$.

Therefore it is a valid operation to multiply the $x$'s into the summations, thus:

Increasing the summation variable $n$ to $n + 2$:

Taking the first 2 terms of the second summation out:

Equating each term to $0$:

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:

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:

Take equation $(3)$:
 * $\ds \sum_{n \mathop = 2}^\infty x^{k - n} \paren {a_{n - 2} \paren {k - n - 2} \paren {k - n + 1} + a_n \paren {p \paren {p + 1} - \paren {k - n} \paren {k - n + 1} } } = 0$

As before, it is assumed that $x^{k - n}$ can never be zero for any value of $k$.

Thus:

Since Legendre's differential equation is a second order ODE, it has two independent solutions.

Solution 1 (for $k = p$)
Thus:

Similarly:

Also:
 * $a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

Solution 2 (for $k = p - 1$)
Thus:

Similarly:

Also:
 * $a_3 = a_5 = a_7 = \dotsb = a_{2 n + 1} = a_1 = 0$

Substituting for $a_n$:

In summary, the two particular solutions are:

and:

Also see
The solutions of Legendre's differential equation are known as Legendre polynomials, which are functions of the parameter $p$.