© examsiri.com
Question : 148 of 160
Marks:
+1,
-0
Solution:
Given,
‌‌y=easin‌−1xOn differentiating w.r.t.
x, we get
‌y1=easin‌−1xa⋅‌‌⇒‌‌y1√1−x2=ay‌⇒‌‌(1−x2)y12=a2y2Again, differentiating w.r.t.
x, we get
‌(1−x2)2y1y2−2xy12=a22yy1⇒‌‌(1−x2)y2−xy1−a2y=0Using 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
© examsiri.com
Go to Question: