$P_n(x)$ is Legendre polynomials and $Q_n(x)$ is Legendre functions of the second kind
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From the general solutions of Legendre polynomials, we know that
\[
P_0(x)=1
\]
However, we can also find the same solution by letting n=0 for the legendre's differential equation
The Legendre differential equation is a second-order ODE, so it has two linearly independent solutions for each n. P_{0}(x)=1 is one of the two solutions for n=0. You found the other solution.
The Legendre differential equation has singular points at [itex]\pm 1[/itex]. For a given n, one of the two solutions will be finite for all finite x, while the other solution will be singular at the singular points [itex]\pm 1[/itex]. The first set of solutions (finite for all finite x) are the Legendre polynomials. The second set are the Legendre functions of the second kind.
The Legendre differential equation is a second-order ODE, so it has two linearly independent solutions for each n. P0(x)=1 is one of the two solutions for n=0. You found the other solution.
The Legendre differential equation has singular points at . For a given n, one of the two solutions will be finite for all finite x, while the other solution will be singular at the singular points . The first set of solutions (finite for all finite x) are the Legendre polynomials. The second set are the Legendre functions of the second kind
BTW, you obviously know LaTeX. This forum supports LaTeX. The start of your post converted to vB LaTeX:
Legendre's differential equation
The general solution is
y = cp(x)1 + dq(x)2 as n= 1,2,3..