$$ y' \left( 1-x^2 \right)+ xy = 0 $$
$$ y' = {-xy \over 1-x^2} $$
$$ {y' \over y} = {-x \over 1-x^2} $$
$$ \int_y^{y_0} {1 \over v} \,dv = {1\over2} \int_x^{x_0} {2u \over u^2-1} \,du $$
$$ \ln |y| -\ln |y_0| = {1\over2} \ln |x^2-1| - {1\over2} \ln |x_0^2-1| $$
$$ \ln |y^2| -\ln |y_0^2| = \ln |x^2-1| - \ln |x_0^2-1| $$
$$ {y^2 \over y_0^2} = {x^2-1 \over x_0^2-1} $$
$$ y^2 = \left( x^2-1 \right) C_0 $$
$$ C_0 = {y_0^2 \over x_0^2-1} $$
Grüße,
M.B.
