I would like to find solutions to the following 2nd order differential equation
$$(1-\mu^{2})\frac{d^{2}g}{d\mu^{2}}+l(l+1)=0,\mu=\cos\theta$$
I think there is a transformation that leads to the Legendre polynomial equation
$$\frac{d}{d\mu}\left[(1-\mu^{2})\frac{dP}{d\mu}\right]+\left[l(l+1)-\frac{m^{2}}{1-\mu^{2}}\right]P=0$$
but I cant figure out which one (I have tried $g=(1-\mu^{2})P$, but that doesnt seem to work).
$$(1-\mu^{2})\frac{d^{2}g}{d\mu^{2}}+l(l+1)=0$$ This equation cannot be transformed into Legendre polynomial equation because it isn't an equation of Legendre kind. Probably there is a typo or something missing.
$$\frac{d^{2}g}{d\mu^2}= -\frac{l(l+1)}{1-\mu^2}=\frac{l(l+1)}{2(1+\mu)}+\frac{l(l+1)}{2(1-\mu)}$$ $$\frac{dg}{d\mu}=\frac{l(l+1)}{2}\ln|1+\mu|-\frac{l(l+1)}{2}\ln|1-\mu|+c_1$$ $$g(x)=\frac{l(l+1)}{2}\left( (1+\mu)\ln|1+\mu|+(1-\mu)\ln|1-\mu| \right)+c_1\mu+c_2$$