The Legendre differential equation is known as:
$ (1-t^2) y'' – 2ty' + n(n+1)y = 0$
I know there is a relation $y(t) = c_1 P_n (t) + c_2 Q_n (t)$, where $P_n (t)$ is the Legendre polynomial.
Given that $P_0 (t) = 1$ is a solution, how can I find $Q_0 (t)$ directly from the diffential equation for $n=0$, where $Q_0 (t)$ is a second independent solution?
(I already know that $Q_0 (t) = \frac{1}{2} \ln(\frac{1+t}{1-t})$)
Best Answer
$ (1-t^2) y'' - 2ty' + n(n+1)y = 0$
case $n=0$ :
$$ (1-t^2) y'' - 2ty' = 0$$
$g(t)=y'$ $$ (1-t^2) g' - 2tg = 0$$ $$\frac{g'}{g}=\frac{2t}{1-t^2}$$ $$g(t)=c_2\frac{1}{1-t^2}$$ $$y(t)=\int g(t)dt=c_2\frac{1}{2} \ln\left| \frac{1+t}{1-t} \right|+c_1$$ $y(t)=c_1P_0(t)+c_2Q_0(t)$ $$P_0(t)=1$$ $$Q_0(t)=\frac{1}{2} \ln\left| \frac{1+t}{1-t} \right|$$