The problem asks for the recurrence relation involving the derivatives of the function $y = e^{m \sin^{-1} x}$. We need to find a relation between $y_{n+2}$, $y_{n+1}$, and $y_n$, where $y_n$ denotes the $n$-th derivative of $y$ with respect to $x$. This is typically solved using the Leibniz differentiation theorem.
Deriving the Recurrence Relation
Step 1: Find the first derivative ($y_1$)
Given $y = e^{m \sin^{-1} x}$.
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{-1} x)$ $y_1 = y \cdot m \cdot \frac{1}{\sqrt{1-x^2}}$Rearranging the terms:
$\sqrt{1-x^2} y_1 = m y$Step 2: Find the second derivative ($y_2$)
Differentiate the equation $\sqrt{1-x^2} y_1 = m y$ with respect to $x$ using the product rule:
$(\frac{d}{dx}\sqrt{1-x^2}) y_1 + \sqrt{1-x^2} (\frac{dy_1}{dx}) = m \frac{dy}{dx}$ $(\frac{-x}{\sqrt{1-x^2}}) y_1 + \sqrt{1-x^2} y_2 = m y_1$Multiply the entire equation by $\sqrt{1-x^2}$:
$-x y_1 + (1-x^2) y_2 = m y_1 \sqrt{1-x^2}$Substitute $\sqrt{1-x^2} y_1 = m y$ from Step 1:
$(1-x^2) y_2 - x y_1 = m (m y)$ $(1-x^2) y_2 - x y_1 = m^2 y$Rearrange to:
$(1-x^2) y_2 - x y_1 - m^2 y = 0$Step 3: Apply Leibniz differentiation theorem
Differentiate the equation $(1-x^2) y_2 - x y_1 - m^2 y = 0$ exactly $n$ times using the Leibniz theorem, which states that $\frac{d^n}{dx^n}(uv) = \sum_{k=0}^{n} \binom{n}{k} (\frac{d^k u}{dx^k}) (\frac{d^{n-k+1} v}{dx^{n-k+1}})$.
Differentiating $(1-x^2) y_2$ $n$ times:
$\frac{d^n}{dx^n}((1-x^2) y_2) = \binom{n}{0} (1-x^2) y_{n+2} + \binom{n}{1} (-2x) y_{n+1} + \binom{n}{2} (-2) y_n$ $= (1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n$Differentiating $-x y_1$ $n$ times:
$\frac{d^n}{dx^n}(-x y_1) = \binom{n}{0} (-x) y_{n+1} + \binom{n}{1} (-1) y_n$ $= -x y_{n+1} - n y_n$Differentiating $-m^2 y$ $n$ times:
$\frac{d^n}{dx^n}(-m^2 y) = -m^2 y_n$Step 4: Combine the results
Summing the results of the differentiation:
$[(1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n] + [-x y_{n+1} - n y_n] - m^2 y_n = 0$Group terms involving $y_{n+2}$, $y_{n+1}$, and $y_n$:
$(1-x^2) y_{n+2} + (-2nx - x) y_{n+1} + (-n(n-1) - n - m^2) y_n = 0$ $(1-x^2) y_{n+2} - (2n+1)x y_{n+1} + (-n^2 + n - n - m^2) y_n = 0$ $(1-x^2) y_{n+2} - (2n+1)x y_{n+1} - (n^2 + m^2) y_n = 0$This derived relation matches option 4.
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