Legendre's differential equation
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\[(1-x^2)y''-2xy'+n(n+1)y=0 \]
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The general solution is
\[y=c_1P_n(x)+c_2Q_n(x) \] where $n=1,2,3,...$
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$P_n(x)$ 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
\[(1-x^2)y''-2xy'+=0 \]
\[\frac{\,dy'}{\,dx}=\frac{2xy'}{1-x^2}\]
\[\int \frac{\,dy'}{y'}=\int \frac{2x}{1-x^2}\]
\[\ln y' = -\ln |1-x^2| + D \]
\[y'=\frac{B}{1-x^2}\]
Find y is pretty straightforward
\[y=A\ln \left[B \left(\frac{1+x}{1-x}\right)\right]\]