Find the Wronskian of two solutions of the given differential equation without solving the
equation.
$$
\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0
$$
$$
\begin{array}{l}{y^{\prime \prime}-\displaystyle\frac{2 x}{1-x^{2}} y^{\prime}+\frac{a(a+1)}{1-x^{2}} y=0} \\ {P(t)=-\displaystyle\frac{2 x}{1-x^{2}}}\end{array}
$$
$$
\begin{aligned} w=c e^{\int-p(t) d t} &=c e^{\displaystyle\int {\frac{2 x}{1-x^{2}} d x}} \qquad u=1-x^{2}, d u=-2 x d x\\ &=\operatorname{ce}^{\displaystyle \int-\frac{1}{u}d u} \\ &=\operatorname{ce}^{-\ln(u)}\\&=\operatorname{ce}^{-\ln \left(1-x^{2}\right)} \\ &=c{(\left.\ln \left(1-x^{2}\right)\right)^{-1}} \\ &=\frac{c}{1-x^{2}} \end{aligned}
$$
