y=ea‌sin−1x On differentiating w.r.t.x, we get y=ea‌sin−1xa.
1
√1−x2
⇒ y1√1−x2=ay ⇒(1−x2)‌y12=a2y2 Again differentiating w.r.t.x, we get (1−x2)‌2‌y1‌y2−2xy12=a22yy1 ⇒ (1−x2)‌y2−xy1−a2y=0 Using Leibnitz's rule, (1−x2)‌yn+2+nC1yn+1(−2x)+nC2yn(−2)−xyn+1−nC1yn−a2yn=0 ⇒ (1−x2)‌yn+2+xyn+1(−2n−1)+yn[−n(n−1)−n−a2]=0 ⇒ (1−x2)‌yn+2−(2n+1)xyn+1=(n2+a2)yn