Question

\( y = e^{a \sin^{-1}(x)} \implies (1 - x^{2}) y_{n+2} - (2n+1)x y_{n+1} \) is equal to

• A. $-(n^2 + a^2)y_n$
• B. $(n^2 - a^2)y_n$
• C. $(n^2 + a^2)y_n$
• D. $-(n^2 - a^2)y_n$

Hint:

Leibnitz Rule helps differentiate products of functions multiple times.

Answer 1

The Correct Option is C

Step 1: First and Second Derivatives
$y_1 = e^{a \sin^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}} \implies y_1 \sqrt{1-x^2} = ay \implies (1-x^2)y_1^2 = a^2 y^2$. Differentiating again: $(1-x^2)2y_1y_2 - 2xy_1^2 = 2a^2yy_1$. Dividing by $2y_1$: $(1-x^2)y_2 - xy_1 - a^2y = 0$.
Step 2: Successive Differentiation (Leibnitz Rule)
Differentiating $n$ times: $[(1-x^2)y_{n+2} + n(-2x)y_{n+1} + \frac{n(n-1)}{2}(-2)y_n] - [xy_{n+1} + n(1)y_n] - a^2y_n = 0$. $(1-x^2)y_{n+2} - xy_{n+1}(2n+1) - y_n[n(n-1) + n + a^2] = 0$.
Step 3: Conclusion
$(1-x^2)y_{n+2} - (2n+1)xy_{n+1} = (n^2 + a^2)y_n$.
Final Answer: (c)