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Write $y(t)={\sum }_{k\ge 0}{a}_k{t}^k$ at the ordinary point zero. Substituting into the Legendre differential equation and comparing coefficients givesThe two arbitrary coefficients $a_0,a_1$ generate the even and odd series separately, convergent at least for $|t|<1$. A nonzero polynomial of degree $n$ has top coefficient satisfying $[\lambda -n(n+1)]a_n=0$, so $\lambda =n(n+1)$ is necessary. For this value the recurrence in parity $n$ terminates at degree $n$, while the other parity cannot terminate unless its initial coefficient is zero. Thus a polynomial solution exists and is unique up to scale, with $\boxed{P_n(-t)=(-1{)}^n{P}_n(t)}$.

For Orthogonality of Legendre polynomials, write the equation in self-adjoint form $[(1-t^2)P_n^{\prime }{]}^{\prime }+n(n+1)P_n=0$. Multiply it by $P_m$, subtract the equation for $P_m$ multiplied by $P_n$, and integrate. The boundary term is $[(1-t^2)(P_m{P}_n^{\prime }-P_n{P}_m^{\prime }){]}_{-1}^1=0$, because the functions are polynomials. For $m\ne n$ the eigenvalues differ, giving ${\int }_{-1}^1{P}_n{P}_m=0$. For $m=n$, the integral of the square is positive. Hence $\int P_n{P}_m=k_n{\delta }_{nm}$ with $k_n>0$.

Let $G(x,t)=(1-2xt+x^2{)}^{-1/2}$. The supplied operator $x{\partial }_x^{2}x$ means $x{\partial }_x^2(xG)$, which equals ${\partial }_x(x^2{\partial }_{x}G)$. Expand $G={\sum }_n{a}_n(x)P_n(t)$ in the given identity and use the Legendre equation. Each coefficient satisfieswhose solutions are $a_n=A_n{x}^n+B_n{x}^{-n-1}$. Regularity at $x=0$ removes $B_n$. At $t=1$, normalization $P_n(1)=1$ and $G(x,1)=1/(1-x)$ give ${\sum }_n{A}_n{x}^n={\sum }_n{x}^n$, so $A_n=1$. ThusA rigorous justification for the expansion and operations is to begin with the Taylor series of $G$ in $x$. It converges uniformly for $|x|\le r<1$, $-1\le t\le 1$. Its $x^n$ coefficient is a degree-$n$ polynomial in $t$, and the differential identity makes that coefficient solve the degree-$n$ Legendre equation. Its value at one is one, so it is exactly $P_n$. This avoids assuming an unjustified endpoint convergence of an arbitrary orthogonal expansion.

Square the generating series and integrate, using uniform convergence and orthogonality:Comparing coefficients proves the generating function and norm of Legendre polynomials result