y''(x)=\frac{y'(x) \left(2 (a-1) x^2-2 a+2 b c \left(x^2-1\right) x^c\right)}{x \left(x^2-1\right)}-\frac{y(x) \left(b c (2 a-c-1) x^{c+2}-b c (2 a-c+1) x^c+x^2 ((a-1) a-v (v+1))-a (a+1)+b^2 c^2 \left(x^2-1\right) x^{2 c}\right)}{x^2 \left(x^2-1\right)}