Determining the coefficient of the legendre polynomial

[Mayan Fung]

We know that the solution to the Legendre equation:
$$ (1-x^2)\frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} + n(n+1) = 0 $$
is the Legendre polynomial $$ y(x) = a_n P_n (x)$$

However, this is a second order differential equation. I am wondering why there is only one leading coefficient. We need two conditions to determine the unique solution to a 2nd order ODE, don't we?

 

[Orodruin]

In addition to the Legendre polynomials, there is a second linearly independent solution to the differential equation, usually denoted $Q_n(x)$. These are usually thrown away as they are singular as $x\to\pm 1$ and a typical requirement is that the functions are regular in these limits.

 

[Mayan Fung]

In addition to the Legendre polynomials, there is a second linearly independent solution to the differential equation, usually denoted $Q_n(x)$. These are usually thrown away as they are singular as $x\to\pm 1$ and a typical requirement is that the functions are regular in these limits.

That solved my puzzle! Thank you!